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

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

[latex]\begin{align}x&=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\[1.5mm] &=\dfrac{-1\pm \sqrt{{1}^{2}-4\cdot 1\cdot \left(2\right)}}{2\cdot 1} \\[1.5mm] &=\dfrac{-1\pm \sqrt{1 - 8}}{2} \\[1.5mm] &=\dfrac{-1\pm \sqrt{-7}}{2} \\[1.5mm] &=\dfrac{-1\pm i\sqrt{7}}{2} \end{align}[/latex]

The solutions to the equation are [latex]x=\dfrac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\dfrac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=-\dfrac{1}{2}+\dfrac{i\sqrt{7}}{2}[/latex] and [latex]x=-\dfrac{1}{2}-\dfrac{i\sqrt{7}}{2}[/latex].

Analysis of the Solution

This quadratic equation has only non-real solutions.

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 [latex]a=2[/latex]. Then we solve for [latex]h[/latex] and [latex]k[/latex].

[latex]\begin{align}&h=-\dfrac{b}{2a}=-\dfrac{4}{2\left(2\right)}=-1\\[2mm]&\end{align}[/latex]

[latex]\begin{align}k&=f\left(h\right)\\&=f\left(-1\right)\\&=2{\left(-1\right)}^{2}+4\left(-1\right)-4\\&=-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{align}&0=2{\left(x+1\right)}^{2}-6 \\ &6=2{\left(x+1\right)}^{2} \\ &3={\left(x+1\right)}^{2} \\ &x+1=\pm \sqrt{3} \\ &x=-1\pm \sqrt{3} \end{align}[/latex]

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

Analysis of the Solution

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

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

[latex]\begin{align}h&=-\dfrac{80}{2\left(-16\right)} \\[1mm]&=\dfrac{80}{32} \\[1mm] &=\dfrac{5}{2} \\[1mm] &=2.5 \end{align}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the [latex]y[/latex] coordinate of the vertex of the parabola.

[latex]\begin{align}k&=H\left(2.5\right) \\[1mm] &=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40 \\[1mm] &=140\hfill \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&=\dfrac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)} \\[1mm] &=\dfrac{-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]t=\dfrac{-80+\sqrt{8960}}{-32}\approx 5.458\hspace{3mm}[/latex] or [latex]\hspace{3mm}t=\dfrac{-80-\sqrt{8960}}{-32}\approx -0.458[/latex]

Since the domain starts at [latex]t=0[/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.

The [latex]x[/latex]-intercepts of the function [latex]f(x)=x^2+2x+3[/latex] are found by setting it equal to zero, and solving for [latex]x[/latex] since the [latex]y[/latex] values of the [latex]x[/latex]-intercepts are zero.

First, identify [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].

[latex]x^2+2x+3=0[/latex]

[latex]a=1,b=2,c=3[/latex]

Substitute these values into the quadratic formula.

[latex]\begin{align}x&=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\\[1mm]&=\dfrac{-2\pm \sqrt{{2}^{2}-4(1)(3)}}{2(1)}\\[1mm]&=\dfrac{-2\pm \sqrt{4-12}}{2} \\[1mm]&=\dfrac{-2\pm \sqrt{-8}}{2}\\[1mm]&=\dfrac{-2\pm 2i\sqrt{2}}{2} \\[1mm]&=-1\pm i\sqrt{2}\\[1mm]x&=-1+i\sqrt{2},-1-i\sqrt{2}\end{align}[/latex]

The solutions to this equation are complex, therefore there are no [latex]x[/latex]-intercepts for the function [latex]f(x)=x^2+2x+3[/latex] in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:

Graph of quadratic function with no [latex]x[/latex]-intercepts in the real numbers.

Note how the graph does not cross the [latex]x[/latex]-axis, therefore there are no real [latex]x[/latex]-intercepts for this function.

Let’s use a diagram such as the one above to record the given information. It is also helpful to introduce a temporary variable, [latex]W[/latex], 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, [latex]W[/latex], in terms of [latex]L[/latex].
   [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]A=LW=L\left(80 - 2L\right)=80L - 2{L}^{2}[/latex]. This formula represents the area of the fence in terms of the variable length [latex]L[/latex]. 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 [latex]a[/latex] is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

To find the vertex:

[latex]h=-\dfrac{80}{2\left(-2\right)}=20[/latex]

and

[latex]\begin{align}k&=A\left(20\right) \\&=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 below.

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, [latex]p[/latex] for price per subscription and [latex]Q[/latex] 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&=\dfrac{79,000 - 84,000}{32 - 30} \\ &=\dfrac{-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 [latex]y[/latex]-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 \\ &\text{Revenue}=p\left(-2,500p+159,000\right) \\ &\text{Revenue}=-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&=-\dfrac{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\hfill \end{align}[/latex]

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.