Answer:
x = 2, x = -6
Step-by-step explanation:
Given quadratic function in vertex form:
[tex]f(x)=(x+2)^2-16[/tex]
The zeros of a function occur when f(x) = 0:
[tex]\implies (x + 2)^2 - 16 = 0[/tex]
Add 16 to both sides:
[tex]\implies (x + 2)^2 - 16+16 = 0+16[/tex]
[tex]\implies (x + 2)^2 =16[/tex]
Square root both sides:
[tex]\implies \sqrt{ (x + 2)^2} =\sqrt{16}[/tex]
[tex]\implies x+2= \pm 4[/tex]
Subtract 2 from both sides:
[tex]\implies x+2-2= -2 \pm 4[/tex]
[tex]\implies x= -2 \pm 4[/tex]
Therefore:
[tex]\implies x=-2+4=2[/tex]
[tex]\implies x=-2-4=-6[/tex]
The zeros of the given quadratic function are x = 2, x = -6.
