Question

Graph the equation y = -x^2+10x - 16 on the accompanying set of axes:

Answer 1

Given the following Quadratic equation:

$y=-x^2+10x-16$

1. You can find the vertex as following:

-Find the x-coordinate of the vertex with this expression:

$\frac{-b}{2a}$

In this case:

$\begin{gathered} a=-1 \\ b=10 \end{gathered}$

Then, substituting values, you get:

$\frac{-10}{2(-1)}=5$

- Substitute this value into the Quadratic equation and evaluate, in order to find the y-coordinate of the vertex. This is:

$\begin{gathered} y=-(5)^2+10(5)-16 \\ y=-25+50-16 \\ y=9 \end{gathered}$

Then, the vertex is:

$(5,9)$

2. Now let's find the roots:

- Substitute the following value of "y" into the equation:

$y=0$

Then:

$0=-x^2+10x-16$

- In order to make the leading coefficient positive, multiply both sides of the equation by -1:

$\begin{gathered} (-1)(0)=(-x^2+10x-16)(-1) \\ 0=x^2-10x+16 \end{gathered}$

- Factor the equation. Find two numbers whose sum is -10 and whose product is 16. These would be -8 and -2. Then:

$(x-8)(x-2)=0$

- The roots are:

$\begin{gathered} x_1=8 \\ x_2=2 \end{gathered}$

3. Now let's find five points to plot them in the coordinate plane. Give five different values to "x" and evaluate in order to find the corresponding value of "y". Then:

- When

$x=1$

You get:

$y=-(1)^2+10(1)-16=-7$

The point is:

$(1,-7)$

- When

$x=3$

You get:

$y=-(3)^2+10(3)-16=5$

The point is:

$(3,5)$

- When:

$x=4$

You get:

$y=-(4)^2+10(4)-16=8$

The point is:

$(4,8)$

- When:

$x=6$$y=-(6)^2+10(6)-16=8$

The point is:

$(6,8)$

- When:

$x=7$

You get:

$y=-(7)^2+10(7)-16=5$

The point is:

$(7,5)$

Finally, you get the following graph of the equation: