Question

How do I find the x-intercepts of a quadratic function in vertex form `f(x)=1/4(x+2)^2-9`?

Answer 1

The x-intercepts of an equation happen at those points when
`f(x) = 0`

Given `f(x) = 1/4(x+2)^2 - 9`
If `f(x) =0`

`1/4(x+2)^2-9 = 0`

`1/4(x+2)^2 = 9`

`(x+2)^2 = 36`

`(x+2) = +-6`

`x= 4` or `x=-8`

Answer 2

Develop to standard form:
`f(x) = 1/4(x^2 + 4x + 2) - 8 = x^2/4 + x + 1 - 9 = x^2 + 4x - 32 = 0`

Find 2 numbers knowing their sum( -b = -4) and their product (c = -32). Compose factor pairs of (-32). Proceed: (-1, 32)(-2, 16)(-4, 8). This last sum is 8 - 4 = 4 = b. Then, the 2 real roots, or x-intercepts are the opposite: 4 and -8.
Answers: x-intercepts : 4 and -8.
Check:
x = 4 -> `f(4) = 36/4 - 9 = 9 - 9 = 0` Correct
x = -8 ->`f(-8) = 36/4 - 9 = 9 = 9 = 0`. Correct.