The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. This parabola does not cross the x-axis, so it has no real zeros. It crosses the y-axis at (0, 7) so this is the y-intercept.

We can see the graph of g is the graph of [latex]f\left(x\right)={x}^{2}[/latex] shifted to the left 2 and down 3, giving a formula in the form [latex]g\left(x\right)=a{\left(x+2\right)}^{2}-3[/latex].

Substituting the coordinates of a point on the curve, such as [latex]\left(0,-1\right)[/latex], we can solve for the stretch factor.

[latex]\begin{align}-1&=a{\left(0+2\right)}^{2}-3 \\ 2&=4a \\ a&=\frac{1}{2} \end{align}[/latex]

In standard form, the algebraic model for this graph is [latex]\left(g\right)x=\frac{1}{2}{\left(x+2\right)}^{2}-3[/latex].

To write this in general polynomial form, we can expand the formula and simplify terms.

[latex]\begin{align}g\left(x\right)&=\frac{1}{2}{\left(x+2\right)}^{2}-3 \\ &=\frac{1}{2}\left(x+2\right)\left(x+2\right)-3 \\ &=\frac{1}{2}\left({x}^{2}+4x+4\right)-3 \\ &=\frac{1}{2}{x}^{2}+2x+2 - 3 \\ &=\frac{1}{2}{x}^{2}+2x - 1 \end{align}[/latex]

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

Analysis of the Solution

We can check our work using the table feature on a graphing utility. First enter [latex]\text{Y1}=\frac{1}{2}{\left(x+2\right)}^{2}-3[/latex]. Next, select [latex]\text{TBLSET,}[/latex] then use [latex]\text{TblStart}=-6[/latex] and [latex]\Delta \text{Tbl = 2,}[/latex] and select [latex]\text{TABLE}\text{.}[/latex]

The ordered pairs in the table correspond to points on the graph.

The horizontal coordinate of the vertex will be at

[latex]\begin{align}h&=-\frac{b}{2a} \\ &=-\frac{-6}{2\left(2\right)} \\ &=\frac{6}{4} \\ &=\frac{3}{2}\end{align}[/latex]

The vertical coordinate of the vertex will be at

[latex]\begin{align}k&=f\left(h\right) \\ &=f\left(\frac{3}{2}\right) \\ &=2{\left(\frac{3}{2}\right)}^{2}-6\left(\frac{3}{2}\right)+7 \\ &=\frac{5}{2} \end{align}[/latex]

Rewriting into standard form, the stretch factor will be the same as the [latex]a[/latex] in the original quadratic.

[latex]\begin{gathered}f\left(x\right)=a{x}^{2}+bx+c \\ f\left(x\right)=2{x}^{2}-6x+7 \end{gathered}[/latex]

Using the vertex to determine the shifts,

[latex]f\left(x\right)=2{\left(x-\frac{3}{2}\right)}^{2}+\frac{5}{2}[/latex]

Analysis of the Solution

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, (k), and where it occurs, (x).

As with any quadratic function, the domain is all real numbers.

Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex.

[latex]\begin{align}h&=-\frac{b}{2a} \\ &=-\frac{9}{2\left(-5\right)} \\ &=\frac{9}{10}& \end{align}[/latex]

The maximum value is given by [latex]f\left(h\right)[/latex].

[latex]\begin{align}f\left(\frac{9}{10}\right)&=5{\left(\frac{9}{10}\right)}^{2}+9\left(\frac{9}{10}\right)-1 \\ &=\frac{61}{20} \end{align}[/latex]

The range is [latex]f\left(x\right)\le \frac{61}{20}[/latex], or [latex]\left(-\infty ,\frac{61}{20}\right][/latex].

Figure 10

Let’s use a diagram such as the one in Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1. We know we have only 80 feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex]. This allows us to represent the width, W, in terms of L.
   [latex]W=80 - 2L[/latex]

   Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

   [latex]\begin{align}A&=LW \\ &=L\left(80 - 2L\right) \\ &=80L - 2{L}^{2} \end{align}[/latex]

   This formula represents the area of the fence in terms of the variable length L. The function, written in general form, is

   [latex]A\left(L\right)=-2{L}^{2}+80L[/latex].
2. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

To find the vertex:

[latex]\begin{align}h&=-\dfrac{80}{2\left(-2\right)} &&& k&=A\left(20\right) \\ &=20 && \text{and} & &=80\left(20\right)-2{\left(20\right)}^{2}\\ &&&&&=800 \end{align}[/latex]

The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

Analysis of the Solution

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11.

Figure 11

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p for price per subscription and Q for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be

[latex]\begin{align}m&=\frac{79,000 - 84,000}{32 - 30} \\ &=\frac{-5,000}{2} \\ &=-2,500 \end{align}[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

[latex]\begin{align}Q&=-2500p+b && \text{Substitute in the point }Q=84,000\text{ and }p=30 \\ 84,000&=-2500\left(30\right)+b && \text{Solve for }b \\ b&=159,000 \end{align}[/latex]

This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers. We now return to our revenue equation.

[latex]\begin{align}\text{Revenue}&=pQ \\ &=p\left(-2,500p+159,000\right) \\ &=-2,500{p}^{2}+159,000p \end{align}[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

[latex]\begin{align}h&=-\frac{159,000}{2\left(-2,500\right)} \\ &=31.8 \end{align}[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

[latex]\begin{align}\text{maximum revenue}&=-2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right) \\ &=2,528,100 \end{align}[/latex]

The maximum revenue is $2,528,100.

Analysis of the Solution

This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.

Figure 12

We begin by solving for when the output will be zero.

[latex]0=2{x}^{2}+4x - 4[/latex]

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

We know that a = 2. Then we solve for h and k.

[latex]\begin{align}h&=-\frac{b}{2a} &&& k&=f\left(-1\right) \\ &=-\frac{4}{2\left(2\right)} &&& &=2{\left(-1\right)}^{2}+4\left(-1\right)-4 \\ &=-1 &&& &=-6 \end{align}[/latex]

So now we can rewrite in standard form.

[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]

We can now solve for when the output will be zero.

[latex]\begin{gathered}2{\left(x+1\right)}^{2}-6=0 \\ 2{\left(x+1\right)}^{2}=6 \\ {\left(x+1\right)}^{2}=3 \\ x+1=\pm \sqrt{3} \\ x=-1\pm \sqrt{3} \end{gathered}[/latex]

The graph has x-intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].

Analysis of the Solution

Figure 15

We can check our work by graphing the given function on a graphing utility and observing the x-intercepts.

Let’s begin by writing the quadratic formula: [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].

When applying the quadratic formula, we identify the coefficients a, b, and c. For the equation [latex]{x}^{2}+x+2=0[/latex], we have a = 1, b = 1, and c = 2. Substituting these values into the formula we have:

[latex]\begin{align}x&=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\ &=\frac{-1\pm \sqrt{{1}^{2}-4\cdot 1\cdot \left(2\right)}}{2\cdot 1} \\ &=\frac{-1\pm \sqrt{1 - 8}}{2} \\ &=\frac{-1\pm \sqrt{-7}}{2} \\ &=\frac{-1\pm i\sqrt{7}}{2} \end{align}[/latex]

The solutions to the equation are [latex]x=\frac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}[/latex]. Note that because of the i, these are non-real zeros.

a. The ball reaches the maximum height at the vertex of the parabola.

[latex]\begin{align} h&=-\frac{80}{2\left(-16\right)} =\frac{80}{32} \\ &=\frac{5}{2} \\ &=2.5 \end{align}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the y coordinate of the vertex of the parabola.

[latex]\begin{align}k&=H\left(-\frac{b}{2a}\right) \\ &=H\left(2.5\right) \\ &=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40 \\ &=140 \end{align}[/latex]

The ball reaches a maximum height of 140 feet.

c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex].

We use the quadratic formula.

[latex]\begin{align} t&=\frac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)} \\ &=\frac{-80\pm \sqrt{8960}}{-32} \end{align}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]\begin{align}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458 && \text{or} && t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458 \end{align}[/latex]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.

Figure 16