Chapter 8, Problem 25CR

To determine

To graph: The quadratic function using zeros.

Explanation of Solution

Given information:

The quadratic function $y=-2x^2+6x+8$

Graph:

Let’s graph using x intercepts and the vertex. Find the x coordinates of vertex using $\frac{-b}{2a}$ where ‘ b’ is coefficient of x term and ‘ a’ is coefficient of $x^2$ term.

So, $a=-2$ and $b=6$

x coordinate of vertex $=\frac{-6}{2(-2)}=1.5$

y coordinate of vertex is given by plug in x as 1.5 into the equation.

  $y=-2{(1.5)}^2+6(1.5)+8$

Simplify it

  $y=-4.5+9+8=12.5$

Vertex is at (1.5, 12.5)

Now, let’s find x intercepts to graph.

x intercept is when $y=0$ :

  $-2x^2+6x+8=0$

Divide both sides by -2. It gives

  $x^2-3x-4=0$

Now, find two integers whose multiplication is -4 but adds to -3 to factor.

It is $-4\times 1=-4$ , $-4+1=-3$ .

So, factored form would be $(x-4)(x+1)=0$

Using zero product property

  $x-4=0$ , $x+1=0$

Solve each equation for ‘ x’

  $\begin{array}{l}x=4\\ x=-1\end{array}$

Let’s represent the vertex (1.5, 12.5) and x intercepts (4,0) and (-1,0) to graph the parabola.

  

Interpretation: The graph is open down parabola with vertex as it’s maximum.