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:

`(-b)/(2a)`

In this case:

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

Then, substituting values, you get:

`(-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: