Identify the coefficients: $$a=2$$, $$b=-8$$, and $$c=7$$.

The $$x$$-coordinate of the vertex will be at

[latex]\begin{align} h &=-\dfrac{b}{2a} \\ h &=-\dfrac{-8}{2\left(2\right)} \\ h &=\dfrac{8}{4} \\ h &=2 \end{align}[/latex]

The $$y$$-coordinate of the vertex will be at

[latex]\begin{align}k &= f(h) \\ k &= f(2) \\ k &= 2(2)^{2}-8(2)+7 \\ k &= 8-16+7\\ k &=-1 \end{align}[/latex]

Rewriting into vertex form, the stretch factor of $$2$$ 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}-8x+7 \end{gathered}[/latex]

Writing in vertex form we found $$a=2$$, $$h=2$$, and $$k=-1$$.

$$f(x)=a(x-h)^2+k$$

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

The vertex is $$(2,-1)$$. Since $$a>0$$, this indicates the parabola opens upward. The minimum value is $$k=-1$$ with the lowest point being $$(2,-1)$$.

Each term has a common factor of [latex]t[/latex], so we can factor out the GCF of $$t$$ and use the zero product principle.

[latex]\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)=0\\t\left(-t+1\right)=0\end{array}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the Zero-Product Property.

[latex]\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{-t}{-1}=\dfrac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}[/latex]

Therefore, [latex]t=0\text{ OR }t=1[/latex].

First, move all the terms to one side. The goal is to try and see if we can use the Zero-Product Property since that is the only tool we currently have for solving quadratic equations.

[latex]\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}[/latex]

We now have all the terms on the left side and zero on the other side.

[latex]\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}[/latex]

We separate our factors into two linear equations using the Zero-Product Property.

[latex]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex]

OR

[latex]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex]

Therefore, [latex]s=-1\text{ OR }s=5[/latex].

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the Zero-Product Property.

$$\begin{align} {x}^{2}-9 &= 0\\ (x-3)(x+3) &= 0 \end{align}$$

$$\begin{array}{cc}\hfill \\ \left(x – 3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=3 & & x=-3 \hfill \end{array}$$

The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the $$b$$ term, [latex]15x[/latex], with two terms using [latex]3[/latex] and [latex]12[/latex] as coefficients of $$x$$. Factor the first two terms, and then factor the last two terms.

Solve using the Zero-Product Property.

[latex]\begin{array}{lll}&\left(4x+3\right)\left(x+3\right)=0 & \hfill \\ \left(4x+3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=-\dfrac{3}{4} & & x=-3 \hfill \end{array}[/latex]

The solutions are [latex]x=-\dfrac{3}{4}[/latex], [latex]x=-3[/latex]. You can see these solutions as the $$x$$-intercepts of the parabola graphed below.

The function can also be written as $$y=2x^2-4x+3$$

Vertex: $$(h,k)$$

$$x$$-coordinate:

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

$$y$$-coordinate:

[latex]\begin{align} k &= f(1) \\ k &= 2(1)^2-4(1)+3 \\ k &= 2-4+3\\ k &=1 \end{align}[/latex]

The vertex is $$(1,1)$$.

Axis of Symmetry is the vertical line that passes through the vertex. That line is $$x=1$$.

$$y$$-intercept: Let $$x=0$$ and solve for $$y$$.

$$\begin{align} y &= 2x^2-4x+3\\ y &= 2(0)^2-4(0)+3\\ y &= 3\end{align}$$

The $$y$$-intercept is $$(0,3)$$.

$$x$$-intercept(s): Let $$y=0$$ and solve for $$x$$. Use factoring to solve.

$$\begin{align} y &= 2x^2-4x+3\\ 0 &= 2x^2-4x+3\end{align}$$

$$a\cdot c = 2\cdot3=6$$ We are looking for two integers that multiply to be $$6$$ and add to be $$-4$$. There are no two integers that satisfy this. This cannot be factored. Later in the book, we will discuss different methods besides factoring in which you can find the $$x$$-intercepts. For now, we will graph this function with the other points we have found and use symmetry of a parabola to find another point on the graph.

We are given that the height of the rocket is [latex]4[/latex] feet from the ground on its way back down. We want to know how long it will take for the rocket to get to that point in its path. We are going to solve for $$t$$.

Substitute [latex]h(t) = 4[/latex] into the formula for height, and try to get zero on one side since we know we can use the Zero-Product Property to solve polynomial equations.

Write and Solve:

[latex]\begin{array}{l}4=-2t^2+7t+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\0=-2t^2+7t\end{array}[/latex]

Now we can factor out a [latex]t[/latex] from each term:

[latex]0=t\left(-2t+7\right)[/latex]

Solve each equation for [latex]t[/latex] using the Zero-Product Property:

[latex]\begin{array}{l}t=0\text{ OR }-2t+7=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-7}\,\,\,\,\,\,\,\underline{-7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{-2t}{-2}=\dfrac{-7}{-2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\dfrac{7}{2}=3.5\end{array}[/latex]

It does not make sense for us to choose [latex]t=0[/latex], because we are interested in the amount of time that has passed when the projectile is [latex]4[/latex] feet from hitting the ground on its way back down. We will choose [latex]t=3.5[/latex].

The rocket will be $$4$$ feet from the ground at [latex]t=3.5\text{ seconds}[/latex].

Figure 4

Let’s use a diagram such as the one in Figure 4 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 does not have a constant term, meaning $$c=0$$. Writing the function in general form we can  identify that [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) \\ h&=20 && \text{and} & k&=-2{\left(20\right)}^{2}+80\left(20\right)\\ &&&&&k=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.

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 5.

Figure 5

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

[latex]\begin{align} h&=-\dfrac{80}{2\left(-16\right)} =\dfrac{80}{32} \\ &=\dfrac{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(-\dfrac{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.