2.1 Quadratic Functions

Vanessa Coffelt

2.1 Quadratic Functions

2.1.1 Graphs of Quadratic Functions

You may recall studying quadratic equations in a previous Algebra course. If not, you may wish to refer to Section 0.5.5 to revisit this topic. In this section, we review those equations in the context of our next family of functions: the quadratic functions.

Definition 2.1

A quadratic function is a function of the form

$f(x) = ax^2 + bx + c,$

where $a$ , $b$ and $c$ are real numbers with $a \neq 0$ .

The domain of a quadratic function is $(-\infty, \infty)$ .

As in Definitions 1.9 and 1.10, the independent variable in Definition 2.1 is $x$ while the values $a$ , $b$ and $c$ are parameters. Note that $a \neq 0$ – otherwise we would have a linear function (see Definition 1.10).

The most basic quadratic function is $f(x) = x^2$ , the squaring function, whose graph appears below along with a corresponding table of values. Its shape may look familiar from your previous studies in Algebra — it is called a parabola. The point $(0,0)$ is called the vertex of the parabola because it is the sole point where the function obtains its extreme value, in this case, a minimum of $0$ when $x = 0$ .

Indeed, the range of $f(x) = x^2$ appears to be $[0, \infty)$ from the graph. We can substantiate this algebraically because for all $x$ , $f(x) = x^2 \geq 0$ . This tells us that the range of $f$ is a subset of $[0, \infty)$ . To show that the range of $f$ actually equals $[0, \infty)$ , we need to show that every real number $c$ in $[0, \infty)$ is in the range of $f$ . That is, for every $c \geq 0$ , we have to show $c$ is an output from $f$ . In other words, we have to show there is a real number $x$ so that $f(x) = x^2 = c$ . Choosing $x = \sqrt{c}$ , we find $f(x) = f(\sqrt{c}) = (\sqrt{c})^2 = c$ , as required.^[1]

A table and a graph side by side. The table has two columns with headers x and f(x) =x^2. The next row is -2 and 4. Then -3/2 and 9/4. The fourth row of the table is -1 and 1. The remaining rows are 0 and 0, 1 and 1, then 3/2 and 9/4, last are 2 and 4. The graph is of the upper half of a coordinate plane. The points from the table are plotted on the graph and connected by a curve opening upward. The curve is symmetric about the y-axis. — Table and Graph of the Basic Quadratic Function

The techniques we used to graph many of the absolute value functions in Section 1.4 can be applied to quadratic functions, too. In fact, knowing the graph of $f(x) = x^2$ enables us to graph every quadratic function, but there’s some extra work involved. We start with the following theorem:

Theorem 2.1

For real numbers $a$ , $h$ and $k$ with $a\neq 0$ , the graph of $F(x) = a(x-h)^2 + k$ is a parabola with vertex $(h,k)$ .

If $a>0$ , the graph resembles ` $\smile$ .’
If $a<0$ , the graph resembles ` $\frown$ .’

Moreover, the vertical line $x=h$ is the axis of symmetry of the graph of $y = F(x)$ .

To prove Theorem 2.1 the reader is encouraged to revisit the discussion following the proof of Theorem 1.4, replacing every occurrence of absolute value notation with the squared exponent.^[2] Alternatively, the reader can skip ahead and read the statement and proof of Theorem 2.2 in Section 2.2. In the meantime we put Theorem 2.1 to good use in the next example.

Example 2.1.1

Example 2.1.1.1a

Graph the following functions using Theorem 2.1. Determine the vertex, zeros and axis-intercepts (if any exist). Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$f(x) = \dfrac{(x-3)^2}{2}$

Solution:

Graph $f(x) = \dfrac{(x-3)^2}{2}$ .

For $f(x) = \frac{(x-3)^2}{2} = \frac{1}{2} (x-3)^2+0$ , we identify $a = \frac{1}{2}$ , $h = 3$ and $k = 0$ .

Thus the vertex is $(3,0)$ and the parabola opens upwards.

The only $x$ -intercept is $(3,0)$ .

Our $y$ -intercept is $\left(0, \frac{9}{2}\right)$ , because $f(0) = \frac{1}{2} (0-3)^2 = \frac{9}{2}$ .

To help us graph the function, it would be nice to have a third point and we’ll use symmetry to find it. The $y$ -value three units to the left of the vertex is $4.5$ , so the $y$ -value must be $4.5$ three units to the right of the vertex as well. Hence, we have our third point: $\left(6, \frac{9}{2}\right)$ .

From the graph, we identify that the range is $[0, \infty)$ and see that $f$ has the minimum value of $0$ at $x = 3$ and no maximum.

Also, $f$ is decreasing on $(-\infty, 3)$ and increasing on $(3, \infty)$ .

The graph is below.

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Example 2.1.1.1b

Graph the following functions using Theorem 2.1. Determine the vertex, zeros and axis-intercepts (if any exist). Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$g(x) = (x+2)^2 - 3$

Solution:

Graph $g(x) = (x+2)^2 - 3$ .

For $g(x) = (x+2)^2 - 3 = (1)(x-(-2))^2+(-3)$ , we identify $a = 1$ , $h = -2$ and $k = -3$ .

This means that the vertex is $(-2,-3)$ and the parabola opens upwards.

Thus we have two $x$ -intercepts. To find them, we set $y = g(x) = 0$ and solve. Doing so yields the equation $(x+2)^2 - 3 = 0$ , or $(x+2)^2 = 3$ . Extracting square roots gives us the two zeros of $g$ : $x + 2 = \pm \sqrt{3}$ , or $x = -2 \pm \sqrt{3}$ . Our $x$ -intercepts are $(-2-\sqrt{3}, 0) \approx (-3.73, 0)$ and $(-2+\sqrt{3}, 0) \approx (-0.27, 0)$ .

We find $g(0) = (0+2)^2-3 = 1$ so our $y$ -intercept is $(0,1)$ .

Using symmetry, we get $(-4,1)$ as another point to help us graph.

The range of $g$ is $[-3, \infty)$ . The minimum of $g$ is $-3$ at $x = -2$ , and $g$ has no maximum.

Moreover, $g$ is decreasing on $(-\infty, -2)$ and $g$ is increasing on $(-2, \infty)$ .

The graph is below.

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Example 2.1.1.1c

Graph the following functions using Theorem 2.1. Determine the vertex, zeros and axis-intercepts (if any exist). Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$h(t) = -2(t-3)^2+1$

Solution:

Graph $h(t) = -2(t-3)^2+1$ .

Given $h(t) = -2(t-3)^2+1$ , we identify $a = -2$ , $h = 3$ and $k = 1$ .

Hence the vertex of the graph is $(3,1)$ and the parabola opens downwards.

Solving $h(t) =-2(t-3)^2+1 = 0$ gives $(t-3)^2 = \frac{1}{2}$ . Extracting square roots\footnote{and rationalizing denominators!} gives $t - 3 = \pm \frac{\sqrt{2}}{2}$ , so that when we add $3$ to each side,^[3] we get $t = \frac{6 \pm \sqrt{2}}{2}$ . Hence, our $t$ -intercepts are $\left(\frac{6 - \sqrt{2}}{2}, 0 \right) \approx (2.29, 0)$ and $\left(\frac{6 + \sqrt{2}}{2}, 0 \right) \approx (3.71, 0)$ .

To find the $y$ -intercept, we compute $h(0) = -2(0-3)^2+1 = -17$ . Thus the $y$ -intercept is $(0,-17)$ .

Using symmetry, we also know $(6,-17)$ is on the graph.

The range of $h$ is $(-\infty, 1]$ . The maximum of $h$ is $1$ at $x = 3$ , and $h$ has no minimum.

Moreover, $h$ is increasing on $(-\infty, 3)$ and $g$ is increasing on $(3, \infty)$ .

The graph is below.

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Example 2.1.1.1d

Graph the following functions using Theorem 2.1. Determine the vertex, zeros and axis-intercepts (if any exist). Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$i(t) = \dfrac{(3-2t)^2 +1}{2}$

Solution:

Graph $i(t) = \dfrac{(3-2t)^2 +1}{2}$ .

We have some work ahead of us to put $i(t)$ into a form we can use to exploit Theorem 2.1:

$\begin{array}{rcl} i(t) &=& \dfrac{(3 - 2t)^2 + 1}{2} \\[8pt] & = & \frac{1}{2} (-2t + 3)^2 + \frac{1}{2} \\[8pt] & = & \frac{1}{2} \left[ -2 \left(t - \frac{3}{2}\right) \right]^2 + \frac{1}{2} \\ [8pt] & = & \frac{1}{2} (-2)^2 \left(t - \frac{3}{2}\right)^2 + \frac{1}{2} \\[8pt] & = & 2\left(t - \frac{3}{2}\right)^2 + \frac{1}{2} \end{array}$

We identify $a = 2$ , $h = \frac{3}{2}$ and $k = \frac{1}{2}$ .

Hence our vertex is $\left(\frac{3}{2}, \frac{1}{2}\right)$ and the parabola opens upwards, meaning there are no $t$ -intercepts.

By computing $i(0) = \frac{(3-2(0))^2 +1}{2} = 5$ , we get $(0,5)$ as the $y$ -intercept.

Using symmetry, this means we also have $(3, 5)$ on the graph.

The range is $\left[ \frac{1}{2}, \infty \right)$ with the minimum of $i$ , $\frac{1}{2}$ , occurring when $t = \frac{3}{2}$ .

Also, $i$ is decreasing on $\left(-\infty, \frac{3}{2} \right)$ and increasing on $\left(\frac{3}{2}, \infty \right)$ .

The graph is given next.

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Example 2.1.1.2

Use Theorem 2.1 to write a possible formula for $H(x)$ whose graph is given below:

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Solution:

Use Theorem 2.1 to write a possible formula for $H(x)$ whose graph is given.

We are instructed to use Theorem 2.1, so we know $H(x) = a(x-h)^2 + k$ for some choice of parameters $a$ , $h$ and $k$ .

The vertex is $(1,3)$ so we know $h = 1$ and $k = 3$ , and hence $H(x) = a(x-1)^2 + 3$ .

To determine the value of $a$ , we use the fact that the $y$ -intercept, as labeled, is $(0,1)$ . This means $H(0) = 1$ , or $a(0-1)^2+ 3 = 1$ . This reduces to $a+3 = 1$ or $a =-2$ .

Our final answer^[4] is $H(x) = -2(x - 1)^2 + 3$ .

A few remarks about Example 2.1.1 are in order. First note that none of the functions are in the form of Definition 2.1. However, if we took the time to perform the indicated operations and simplify, we’d determine:

$f(x) = \frac{(x-3)^2}{2} = \frac{1}{2} x^2 - 3x + \frac{9}{2}$
$h(t) = -2(t-3)^2+1 = -2t^2+12t-17$
$g(x) = (x+2)^2 - 3 = x^2+4x+1$
$i(t) = \frac{(3-2t)^2 +1}{2} = 2t^2-6t+5$

While the $y$ -intercepts of the graphs of the each of the functions are easier to see when the formulas for the functions are written in the form of Definition 2.1, the vertex is not. For this reason, the form of the functions presented in Theorem 2.1 are given a special name.

Definition 2.2 Standard and General Form of Quadratic Functions

The general form of the quadratic function $f$ is $f(x) = ax^2+bx+c$ , where $a$ , $b$ and $c$ are real numbers with $a \neq 0$ .
The standard form of the quadratic function $f$ is $f(x) = a(x-h)^2 + k$ , where $a$ , $h$ and $k$ are real numbers with $a\neq 0$ . The standard form is often called the vertex form.

If we proceed as in the remarks following Example 2.1.1, we can convert any quadratic function given to us in standard form and convert to general form by performing the indicated operation and simplifying:

$\begin{array}{rcl} f(x) & = & a(x-h)^2 + k \\ &= & a \left(x^2 -2hx + h^2 \right) + k \\ & = & ax^2 - 2ahx + ah^2 + k \\ & = & a x^2 + (-2ah)x + (ah^2+k). \\ \end{array}$

With the identifications $b = -2ah$ and $c = ah^2+k$ , we have written $f(x)$ in the form $f(x) = ax^2 + bx+c$ . Likewise, through a process known as `completing the square’, we can take any quadratic function written in general form and rewrite it in standard form. We briefly review this technique in the following example — for a more thorough review the reader should see Section 0.5.5.

Example 2.1.2

Example 2.1.2.1

Graph the following functions. Determine the vertex, zeros and axis-intercepts, if any exist. Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$f(x) = x^2-4x+3$

Solution:

Graph $f(x) = x^2-4x+3$ . Determine the vertex, zeros and axis-intercepts, if any exist. Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

We follow the procedure for completing the square in Section 0.5.5. The only difference here is instead of the quadratic equation being set to $0$ , it is equal to $f(x)$ . This means when we are finished completing the square, we need to solve for $f(x)$ .

$\begin{array}{rclr} f(x) & = & x^2 - 4x+3 & \\ [4pt] f(x) - 3 & = & x^2-4x & \text{Subtract $3$ from both sides.} \\ [4pt] f(x) - 3 + (-2)^2 & = & x^2-4x+(-2)^2 & \text{Add $\left(\frac{1}{2}(-4)\right)^2$ to both sides.} \\ [4pt] f(x) + 1 & = & (x-2)^2 & \text{Factor the perfect square trinomial.} \\ [4pt] f(x) & = & (x-2)^2 - 1 & \text{Solve for $f(x)$.} \\ \end{array}$

The reader is encouraged to start with $f(x) = (x-2)^2-1$ , perform the indicated operations and simplify the result to $f(x) = x^2-4x+3$ .

From the standard form, $f(x) = (x-2)^2-1$ , we see that the vertex is $(2,1)$ and that the parabola opens upwards.

To find the zeros of $f$ , we set $f(x) = 0$ .

We have two equivalent expressions for $f(x)$ so we could use either the general form or standard form. We solve the former and leave it to the reader to solve the latter to see that we get the same results either way. To solve $x^2 - 4x + 3 = 0$ , we factor: $(x-3)(x-1) = 0$ and obtain $x = 1$ and $x =3$ . We get two $x$ -intercepts, $(1,0)$ and $(3,0)$ .

To find the $y$ -intercept, we need $f(0)$ . Again, we could use either form of $f(x)$ for this and we choose the general form and find that the $y$ -intercept is $(0,3)$ .

From symmetry, we know the point $(4,3)$ is also on the graph.

We see that the range of $f$ is $[-1,\infty)$ with the minimum $-1$ at $x = 2$ .

Finally, $f$ is decreasing on $(-\infty, 2)$ and increasing from $(2, \infty)$ .

The graph is below.

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Example 2.1.2.2

Graph the following functions. Determine the vertex, zeros and axis-intercepts, if any exist. Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

$g(t) = 6 - 4t -2t^2$

Solution:

Graph $g(t) = 6 - 4t -2t^2$ . Determine the vertex, zeros and axis-intercepts, if any exist. Identify the extrema and then list the intervals over which the function is increasing, decreasing or constant.

We first rewrite $g(t) = 6 - 4t - 2t^2$ as $g(t) = -2t^2 - 4t + 6$ . As with the previous example, once we complete the square, we solve for $g(t)$ :

$\begin{array}{rclr} g(t) & = & -2t^2-4t+6 & \\ [6pt] g(t) - 6 & = & -2t^2-4t & \text{Subtract $6$ from both sides.} \\ [6pt] \dfrac{g(t) - 6}{-2} & = & \dfrac{ -2t^2-4t }{-2} & \text{Divide both sides by $-2$.}\\ [10pt] \dfrac{g(t) - 6}{-2} + (1)^2 & = & t^2+2t +(1)^2 & \text{Add $\left( \frac{1}{2} (2) \right)^2$ to both sides.} \\ [10pt] \dfrac{g(t) - 6}{-2} + 1 & = & (t+1)^2 & \text{Factor the prefect square trinomial.} \\ [10pt] \dfrac{g(t) - 6}{-2} & = & (t+1)^2 - 1 & \\ [10pt] g(t) - 6 & = & -2 \left[ (t+1)^2-1 \right] & \\ [6pt] g(t) & = & -2(t+1)^2 + 2 + 6 & \\ [6pt] g(t) & = & -2(t+1)^2+8 \\ \end{array}$

We can check our answer by expanding $-2(t+1)^2+8$ and show that it simplifies to $-2t^2 - 4t+6$ .

From the standard form, we identify the vertex is $(-1,8)$ and that the parabola opens downwards.

Setting $g(t) = -2t^2 - 4t+6= 0$ , we factor to get $-2(t-1)(t+3) = 0$ so $t = -3$ and $t = 1$ . Hence, our two $t$ -intercepts are $(-3,0)$ and $(1,0)$ .

The $y$ -intercept is $(0,6)$ , as $g(0) = 6$ .

Using symmetry, we also have the point $(-2,6)$ on the graph.

The range is $(-\infty, 8]$ with a maximum of $8$ when $t = -1$ .

Finally we note that $g$ is increasing on $(-\infty, -1)$ and decreasing on $(-1, \infty)$ .

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We now generalize the procedure demonstrated in Example 2.1.2. Let $f(x) = ax^2 + bx + c$ for $a \neq 0$ :

$\begin{array}{rclr} f(x) & = & ax^2 + bx +c \\ [5pt] f(x) - c & = & ax^2 + bx & \text{Subtract $c$ from both sides.}\\ [5pt] \dfrac{f(x)-c}{a} & = & \dfrac{ax^2 + bx}{a} & \text{Divide both sides by $a \neq 0$.} \\ [10pt] \dfrac{f(x)-c}{a} & = & x^2 + \dfrac{b}{a} x & \\ [10pt] \dfrac{f(x)-c}{a} + \left(\dfrac{b}{2a}\right)^2 & = & x^2 + \dfrac{b}{a} x + \left(\dfrac{b}{2a}\right)^2 & \text{Add $ \left(\dfrac{b}{2a}\right)^2 $ to both sides.}\\ [10pt] \dfrac{f(x)-c}{a} + \dfrac{b^2}{4a^2} & = & \left(x + \dfrac{b}{2a}\right)^2 & \text{Factor the perfect square trinomial.} \\ [10pt] \dfrac{f(x)-c}{a} & = & \left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} & \text{Solve for $f(x)$.}\\ [10pt] f(x)-c & = & a \left[ \left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2}\right] & \\ [10pt] f(x)-c & = & a\left(x + \dfrac{b}{2a}\right)^2 - a\dfrac{b^2}{4a^2} & \\ [10pt] f(x) & = & a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c & \\ [10pt] f(x) & = & a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a} & \text{Get a common denominator.} \\ \end{array}$

By setting $h = -\frac{b}{2a}$ and $k = \frac{4ac - b^2}{4a}$ , we have written the function in the form $f(x) = a(x-h)^2 + k$ . This establishes the fact that every quadratic function can be written in standard form. Moreover, writing a quadratic function in standard form allows us to identify the vertex rather quickly, and so our work also shows us that the vertex of $f(x) = ax^2+bx+c$ is $\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)$ . It is not worth memorizing the expression $\frac{4ac - b^2}{4a}$ due to the fact that we can write this as $f\left(-\frac{b}{2a}\right)$ .

We summarize the information detailed above in the following box:

Equation 2.1 Vertex Formulas for Quadratic Functions

Suppose $a$ , $b$ , $c$ , $h$ and $k$ are real numbers where $a \neq 0$ .

If $f(x) = a(x-h)^2 + k$ , then the vertex of the graph of $y=f(x)$ is the point $(h,k)$ .
If $f(x) = ax^2+bx+c$ , then the vertex of the graph of $y=f(x)$ is the point $\left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right)$ .

Completing the square is also the means by which we may derive the celebrated Quadratic Formula, a formula which returns the solutions to $ax^2+bx+c = 0$ for $a \neq 0$ . Before we state it here for reference, we wish to encourage the reader to pause a moment and read the derivation if the Quadratic Formula found in Section 0.5.5. The work presented in this section transforms the general form of a quadratic function into the standard form whereas the work in Section 0.5.5 finds a formula to solve an equation. There is great value in understanding the similarities and differences between the two approaches.

Equation 2.2 The Quadratic Formula

The zeros of the quadratic function $f(x) = ax^2+bx+c$ are:

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

It is worth pointing out the symmetry inherent in Equation 2.2. We may rewrite the zeros as:

$\begin{array}{rcl} x &=& \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\[8pt] &=& -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}, \end{array}$

so that, if there are real zeros, they (like the rest of the parabola) are symmetric about the line $x = -\frac{b}{2a}$ . Another way to view this symmetry is that the $x$ -coordinate of the vertex is the average of the zeros. We encourage the reader to verify this fact in all of the preceding examples, where applicable.

Next, recall that if the quantity $b^2-4ac$ is strictly negative then we do not have any real zeros. This quantity is called the discriminant and is useful in determining the number and nature of solutions to a quadratic equation. We remind the reader of this below.

Equation 2.3 The Discriminant of a Quadratic Function

Given a quadratic function in general form $f(x) = ax^2 + bx + c$ , the discriminant is the quantity $b^2-4ac$ .

If $b^2-4ac>0$ then $f$ has two unequal (distinct) real zeros.
If $b^2-4ac=0$ then $f$ has one (repeated) real zero.
If If $b^2-4ac<0$ then $f$ has two unequal (distinct) non-real zeros.

We’ll talk more about what we mean by a `repeated’ zero and how to compute `non-real’ zeros in Chapter 2. For us, the discriminant has the graphical implication that if $b^2-4ac>0$ then we have two $x$ -intercepts; if $b^2-4ac=0$ then we have just one $x$ -intercept, namely, the vertex; and if $b^2-4ac<0$ then we have no $x$ -intercepts because the parabola lies entirely above or below the $x$ -axis. We sketch each of these scenarios below assuming $a>0$ . (The sketches for $a<0$ are similar.)

Three side by side graphs. The first graph includes a parabola with two x intercepts, and shows the axis of symmetry at x = -b/2a. The graph is labeled with the discriminant greater than zero. The second graph includes a parabola with one x intercept, at the vertex, and shows the axis of symmetry at x = -b/2a. The graph is labeled with the discriminant equals than zero. The first graph includes a parabola completely above the horizontal axis with no x intercepts, and shows the axis of symmetry at x = -b/2a. The graph is labeled with the discriminant less than zero. — Graphical Representations of the 3 Cases for a Discriminant of a Quadratic Function

We now revisit the economic scenario first described in Examples 1.3.8 and 1.3.9 where we were producing and selling PortaBoy game systems. Recall that the cost to produce $x$ PortaBoys is denoted by $C(x)$ and the price-demand function, that is, the price to charge in order to sell $x$ systems is denoted by $p(x)$ . We introduce two more related functions below: the revenue and profit functions.

Definition 2.3 Revenue and Profit

Suppose $C(x)$ represents the cost to produce $x$ units and $p(x)$ is the associated price-demand function. Under the assumption that we are producing the same number of units as are being sold:

The revenue obtained by selling $x$ units is $R(x) = x \, p(x)$ .
That is, revenue = (number of items sold) $\cdot$ (price per item).
The profit made by selling $x$ units is $P(x) = R(x) - C(x)$ .
That is, profit = (revenue) – (cost).

Said differently, the revenue is the amount of money collected by selling $x$ items whereas the profit is how much money is left over after the costs are paid.

Example 2.1.3

Example 2.1.3.1

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Write formulas for the associated revenue and profit functions; include the domain of each.

Solution:

Write formulas for the associated revenue and profit functions, including the domain of each.

The formula for the revenue function is

$\begin{array}{rcl} R(x) &=& x \, p(x) \\ &=& x(-1.5x+250) \\ &=& -1.5x^2 + 250x. \end{array}$

The domain of $p$ is restricted to $0 \leq x \leq 166$ , and thus the domain of $R$ .

To find the profit function $P(x)$ , we subtract

$\begin{array}{rcl} P(x) &=& R(x) - C(x) \\ &=& \left(-1.5x^2+250x\right) - \left(80x + 150\right) \\ &=& -1.5x^2+170x-150. \end{array}$

The cost function formula is valid for $x \geq 0$ , but the revenue function is valid when $0 \leq x \leq 166$ . Hence, the domain of $P$ is likewise restricted to $[0, 166]$ .

Example 2.1.3.2

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Compute and interpret $P(0)$ .

Solution:

Compute and interpret $P(0)$ .

We find

$\begin{array}{rcl} P(0) &=& -1.5(0)^2+170(0) - 150 \\ &=& -150. \end{array}$

This means that if we produce and sell 0 PortaBoy game systems, we have a profit of $-$ 150 dollars.

As profit = (revenue) $-$ (cost), this means our costs exceed our revenue by 150 dollars. This makes perfect sense, we don’t sell any systems and our revenue is 0 dollars, but our fixed costs (see Example 1.3.8) are 150 dollars.

Example 2.1.3.3

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Compute and interpret the zeros of $P$ .

Solution:

Compute and interpret the zeros of $P$ .

To find the zeros of $P$ , we set $P(x) = 0$ and solve $-1.5x^2+170x-150=0$ . Factoring here would be challenging to say the least, so we use the Quadratic Formula, Equation 2.2. Identifying $a = -1.5$ , $b=170$ and $c=-150$ , we obtain

$\begin{array}{rclr} x & = & \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} & \\ [10pt] & = & \dfrac{-170 \pm \sqrt{170^2 - 4(-1.5)(-150)}}{2(-1.5)} & \\ [10pt] \end{array}$

$\begin{array}{rclr} x & = & \dfrac{-170 \pm \sqrt{28000}}{-3} & \\ [10pt] & = & \dfrac{170 \pm 20 \sqrt{70}}{3} & \\ [10pt] & \approx & 0.89, \, 112.44. \\ & & \phantom{\dfrac{-170 \pm \sqrt{170^2 - 4(-1.5)(-150)}}{2(-1.5)} }& \end{array}$

Given that profit = (revenue) $-$ (cost), if profit = 0, then revenue = cost. Hence, the zeros of $P$ are called the `break-even’ points – where just enough product is sold to recover the cost spent to make the product. Also, $x$ represents a number of game systems, which is a whole number, so instead of using the exact values of the zeros, or even their approximations, we consider $x = 0$ and $x = 1$ along with $x = 112$ and $x=113$ .

We find $P(0) = -150$ , $P(1) = 18.5$ , $P(112) = 74$ and $P(113)= -93.5$ .

These data suggest that, in order to be profitable, at least 1 but not more than 112 systems should be produced and sold, as borne out in number 4.

Example 2.1.3.4

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Graph $y = P(x)$ . Find the vertex and axis intercepts.

Solution:

Graph $y = P(x)$ . Find the vertex and axis intercepts.

Knowing the zeros of $P$ , we have two $x$ -intercepts: $\left( \frac{170 - 20 \sqrt{70}}{3},0\right) \approx (0.89,0)$ and $\left( \frac{170 + 20 \sqrt{70}}{3},0\right) \approx (112.44,0)$ .

As $P(0)=-150$ , we get the $y$ -intercept is $(0,-150)$ .

To find the vertex, we appeal to Equation 2.1. Substituting $a = -1.5$ and $b=170$ , we get

$\begin{array}{rcl} x &=& -\frac{170}{2(-1.5)} \\[6pt] &=& \frac{170}{3} \\[6pt] &=& 56.\overline{6} \end{array}$

To find the $y$ -coordinate of the vertex, we compute

$\begin{array}{rcl} P\left( \frac{170}{3} \right) &=& \frac{14000}{3} \\[6pt] &=& 4666.\overline{6} \end{array}$

Hence, the vertex is $(56.\overline{6}, 4666.\overline{6})$ .

The domain is restricted $0 \leq x \leq 166$ and we find $P(166) = -13264$ . Attempting to plot all of these points on the same graph to any sort of scale is challenging. Instead, we present a portion of the graph for $0 \leq x \leq 113$ . Even with this, the intercepts near the origin are crowded.

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Example 2.1.3.5

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Interpret the vertex of the graph of $y = P(x)$ .

Solution:

Interpret the vertex of the graph of $y = P(x)$ .

From the vertex, we see that the maximum of $P$ is $4666.\overline{6}$ when $x = 56.\overline{6}$ . As before, $x$ represents the number of PortaBoy systems produced and sold, so we cannot produce and sell $56.\overline{6}$ systems.

Hence, by comparing $P(56) = 4666$ and $P(57)=4666.5$ , we conclude that we will make a maximum profit of 4666.50 dollars if we sell 57 game systems.

Example 2.1.3.6

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

What should the price per system be in order to maximize profit?

Solution:

What should the price per system be in order to maximize profit?

We’ve determined that we need to sell 57 PortaBoys to maximize profit, so we substitute $x=57$ into the price-demand function to get $p(57) = -1.5(57)+250 = 164.5$ .

In other words, to sell 57 systems, and thereby maximize the profit, we should set the price at 164.50 dollars per system.

Example 2.1.3.7

In Example 1.3.8 the cost to produce $x$ PortaBoy game systems for a local retailer was given by $C(x) = 80x + 150$ for $x \geq 0$ and in Example 1.3.9 the price-demand function was found to be $p(x) = -1.5x+250$ , for $0 \leq x \leq 166$ .

Compute and interpret the average rate of change of $P$ over the interval $[0, 57]$ .

Solution:

Compute and interpret the average rate of change of $P$ over the interval $[0, 57]$ .

To find the average rate of change of $P$ over $[0, 57]$ , we compute

$\begin{array}{rcl} \dfrac{\Delta [P(x)]}{\Delta x} &=& \dfrac{P(57) - P(0)}{57-0} \\[8pt] &=& \dfrac{4666.5 - (-150)}{57} \\[8pt] &=& 84.5 \end{array}$

This means that as the number of systems produced and sold ranges from 0 to 57, the average profit per system is increasing at a rate of 84.50 dollars. In other words, for each additional system produced and sold, the profit increased by 84.50 dollars on average.

We hope Example 2.1.3 shows the value of using a continuous model to describe a discrete situation. True, we could have `run the numbers’ and computed $P(1), \, P(2), \, \ldots, \, P(166)$ to eventually determine the maximum profit, but the vertex formula made much quicker work of the problem.

Our next example is classic application of optimizing a quadratic function.

Example 2.1.4

Example 2.1.3.7

Much to Donnie’s surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives so the time is right for him to pursue his dream of raising alpaca. He wishes to build a rectangular pasture and estimates that he has enough money for 200 linear feet of fencing material. If he makes the pasture adjacent to a river (so that no fencing is required on that side), what are the dimensions of the pasture which maximize the area? What is the maximum area? If an average alpaca needs 25 square feet of grazing area, how many alpaca can Donnie keep in his pasture?

Solution:

We are asked to determine the dimensions of the pasture which would give a maximum area, so we begin by sketching the diagram seen below.

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We let $w$ denote the width of the pasture and we let $\ell$ denote the length of the pasture. The units given to us in the statement of the problem are feet, so we assume that $w$ and $\ell$ are measured in feet. The area of the pasture, which we’ll call $A$ , is related to $w$ and $\ell$ by the equation $A = w \ell$ . Given $w$ and $\ell$ are both measured in feet, $A$ has units of $\text{feet}^2$ , or square feet.

We are also told that the total amount of fencing available is $200$ feet, which means $w + \ell + w = 200$ , or, $\ell+2w = 200$ .

We now have two equations, $A = w \ell$ and $\ell+2w = 200$ .

In order to use the tools given to us in this section to maximize $A$ , we need to use the information given to write $A$ as a function of just one variable, either $w$ or $\ell$ . This is where we use the equation $\ell+2w = 200$ . Solving for $\ell$ , we find $\ell = 200-2w$ , and we substitute this into our equation for $A$ . We get

$\begin{array}{rcl} A &=& w \ell \\ &=& w(200-2w) \\ &=& 200w-2w^2 \end{array}$

We now have $A$ as a function of $w$ ,

$\begin{array}{rcl} A &=& f(w) \\ &=& 200w-2w^2 \\ &=& -2w^2+200w \end{array}$

Before we go any further, we need to find the applied domain of $f$ so that we know what values of $w$ make sense in this situation.^[5] Given that $w$ represents the width of the pasture we need $w > 0$ .

Likewise, $\ell$ represents the length of the pasture, so $\ell = 200-2w > 0$ . Solving this latter inequality yields $w < 100$ .

Hence, the function we wish to maximize is $f(w) = -2w^2 + 200w$ for $0 < w < 100$ . We know two things about the quadratic function $f$ : the graph of $A = f(w)$ is a parabola and (with the coefficient of $w^2$ being $-2$ ) the parabola opens downwards.

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This means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we find

$w = -\frac{200}{2(-2)} = 50$

and

$\begin{array}{rcl} A &=& f(50) \\ &=& -2(50)^2 + 200(50)\\ &=& 5000 \end{array}$

$w=50$ lies in the applied domain, $0 < w < 100$ , thus we have that the area of the pasture is maximized when the width is $50$ feet.

To find the length, we use $\ell = 200-2w$ and find $\ell = 200-2(50) = 100$ , so the length of the pasture is $100$ feet.

The maximum area is $A =f(50) = 5000$ , or $5000$ square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise $\frac{5000}{25} = 200$ average alpaca.

The function $f$ in Example 2.1.4 is called the objective function for this problem – it’s the function we’re trying to optimize. In the case above, we were trying to maximize $f$ . The equation $\ell+2w = 200$ along with the inequalities $w>0$ and $\ell >0$ are called the constraints. As we saw in this example, and as we’ll see again and again, the constraint equation is used to rewrite the objective function in terms of just one of the variables where constraint inequalities, if any, help determine the applied domain.

2.1.2 Section Exercises

In Exercises 1 – 9, graph the quadratic function. Determine the vertex and axis intercepts of each graph, if they exist. State the domain and range, identify the maximum or minimum, and list the intervals over which the function is increasing or decreasing. If the function is given in general form, convert it into standard form; if it is given in standard form, convert it into general form.

$f(x) = x^{2} + 2$
$f(x) = -(x + 2)^{2}$
$f(x) = x^{2} - 2x - 8$
$g(t) = -2(t + 1)^{2} + 4$
$g(t) = 2t^2 - 4t - 1$
$g(t) = -3t^{2} + 4t - 7$
$h(s) = s^2 + s + 1$
$h(s) = -3s^2+5s+4$
$h(s) = s^{2} - \dfrac{1}{100} s - 1$

In Exercises 10 – 13, write a formula for each function below in the form $F(x) = a(x-h)^2+k$ .

In Exercises 14 – 18, cost and price-demand functions are given. For each scenario,

Determine the profit function $P(x)$ .
Compute the number of items which need to be sold in order to maximize profit.
Calculate the maximum profit.
Calculate the price to charge per item in order to maximize profit.
Identify and interpret break-even points.

The cost, in dollars, to produce $x$ “I’d rather be a Sasquatch” T-Shirts is $C(x) = 2x+26$ , $x \geq 0$ and the price-demand function, in dollars per shirt, is $p(x) = 30 - 2x$ , for $0 \leq x \leq 15$ .
The cost, in dollars, to produce $x$ bottles of $100 \%$ All-Natural Certified Free-Trade Organic Sasquatch Tonic is $C(x) = 10x+100$ , $x \geq 0$ and the price-demand function, in dollars per bottle, is $p(x) = 35 - x$ , for $0 \leq x \leq 35$ .
The cost, in cents, to produce $x$ cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is $C(x) = 18x + 240$ , $x \geq 0$ and the price-demand function, in cents per cup, is $p(x) = 90-3x$ , for $0 \leq x \leq 30$ .
The daily cost, in dollars, to produce $x$ Sasquatch Berry Pies is $C(x) = 3x + 36$ , $x \geq 0$ and the price-demand function, in dollars per pie, is $p(x) = 12-0.5x$ , for $0 \leq x \leq 24$ .
The monthly cost, in hundreds of dollars, to produce $x$ custom built electric scooters is $C(x) = 20x + 1000$ , $x \geq 0$ and the price-demand function, in hundreds of dollars per scooter, is $p(x) = 140-2x$ , for $0 \leq x \leq 70$ .
The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking $x$ cookies is $C(x) = 0.1x + 25$ and that the demand function for their cookies is $p = 10 - .01x$ for $0 \leq x \leq 1000$ . How many cookies should they bake in order to maximize their profit?
Using data from Bureau of Transportation Statistics, the average fuel economy $F(t)$ in miles per gallon for passenger cars in the US $t$ years after 1980 can be modeled by $F(t) = -0.0076t^2+0.45t + 16$ , $0 \leq t \leq 28$ . Find and interpret the coordinates of the vertex of the graph of $y = F(t)$ .
The temperature $T$ , in degrees Fahrenheit, $t$ hours after 6 AM is given by:
$T(t) = -\frac{1}{2} t^2 + 8t+32, \quad 0 \leq t \leq 12$

What is the warmest temperature of the day? When does this happen?
Suppose $C(x) = x^2-10x+27$ represents the costs, in hundreds, to produce $x$ thousand pens. How many pens should be produced to minimize the cost? What is this minimum cost?
Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Due to the fact that one side of the garden will border the house, Skippy doesn’t need fencing along that side. What are the dimensions of the garden which will maximize the area of the garden? What is the maximum area of the garden?
In the situation of Example 2.1.4, Donnie has a nightmare that one of his alpaca fell into the river. To avoid this, he wants to move his rectangular pasture away from the river so that all four sides of the pasture require fencing. If the total amount of fencing available is still 200 linear feet, what dimensions maximize the area of the pasture now? What is the maximum area? Assuming an average alpaca requires 25 square feet of pasture, how many alpaca can he raise now?
What is the largest rectangular area one can enclose with 14 inches of string?
The two towers of a suspension bridge are 400 feet apart. The parabolic cable^[6] attached to the tops of the towers is 10 feet above the point on the bridge deck that is midway between the towers. If the towers are 100 feet tall, find the height of the cable directly above a point of the bridge deck that is 50 feet to the right of the left-hand tower.

In Exercises 27 – 32, solve the quadratic equation for the indicated variable.

$x^{2} - 10y^{2} = 0$ for $x$
$y^{2} - 4y = x^{2} - 4$ for $x$
$x^{2} - mx = 1$ for $x$
$y^{2} - 3y = 4x$ for $y$
$y^{2} - 4y = x^{2} - 4$ for $y$
$-gt^{2} + v_{0}t + s_{0} = 0$ for $t$ (Assume $g \neq 0$ .)
(This is a follow-up to Exercise 96 in Section 1.3.1.) The Lagrange Interpolate function for three points , , and where , , and are three distinct real numbers is given by:
$L(x) = y_{0} \dfrac{(x - x_{1}) (x - x_{2}) }{(x_{0} - x_{1})(x_{0} - x_{2})}+ y_{1} \dfrac{(x - x_{0}) (x - x_{2}) }{(x_{1} - x_{0})(x_{1} - x_{2})} + y_{2} \dfrac{(x - x_{0}) (x - x_{1}) }{(x_{2} - x_{0})(x_{2} - x_{1})}$
1. For each of the following sets of points, find using the formula above and verify each of the points lies on the graph of .
  1. $(-1,1)$ , $(1,1)$ , $(2,4)$
  2. $(1,3)$ , $(2,10)$ , $(3,21)$
  3. $(0,1)$ , $(1,5)$ , $(2,7)$
2. Verify that, in general, $L(x_{0}) = y_{0}$ , $L(x_{1}) = y_{1}$ , and $L(x_{2}) = y_{2}$ .
3. Find $L(x)$ for the points $(-1, 6)$ , $(1, 4)$ and $(3,2)$ . What happens?
4. Under what conditions will $L(x)$ produce a quadratic function? Make a conjecture, test some cases, and prove your answer.

Section 2.1 Exercise Answers can be found in the Appendix … Coming soon

This assumes, of course, $\sqrt{c}$ is a real number for all real numbers $c \geq 0$ \ldots ↵
i.e., replace $|x|$ with $x^2$ , $|c|$ with $c^2$ , $|x-h|$ with $(x-h)^2$ . ↵
and get common denominators! ↵
The reader is encouraged to compare this example with number 2 of Example 1.4.2. ↵
Donnie would be very upset if, for example, we told him the width of the pasture needs to be $-50$ feet. ↵
The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in an exercise in Section 5.7 what shape a free hanging cable makes. ↵