Question

Find the X intercept and coordinates of the vertex for the parabola Y=X^2+8X +16 . if there is more than one x-intercept, separate them with commas.

Answer 1

Given an equation of the parabola:

`y=x^2+8x+16`

We have to find the vertex and the x-intercept of the parabola.

It is known that the vertex of the parabola is equal to -b/2a.

Here, a = 1, b = 8, c = 16. So,

`\begin{gathered} x=(-8)/(2(1)) \\ x=(-8)/(2) \\ x=-4 \end{gathered}`

At x = -4, the y-coordinate is:

`\begin{gathered} y=(-4)^2+8(-4)+16 \\ y=16-32+16 \\ y=0 \end{gathered}`

Thus, the vertex is (-4, 0).

Now, we know that the x-intercept can be obtained by setting y = 0.

`\begin{gathered} y=0 \\ x^2+8x+16=0 \\ (x+4)^2=0 \\ x+4=0 \\ x=-4 \end{gathered}`

Thus, there is only one x-intercept -4.