A vertical parabola has vertex (2,-8) and y intercept 4. What are its x intercepts?

Answer 1

ax^2 + bx + c = 0 ... vertex = -b / 2a =2
c = 4 (intercept value)
if a = 2, then b = -4
so
2x^2 -4x + 4....... vertex = -(-4) / 2 .. =2 check
2(0)^2 - 4(0) + 4 = 4 (0,4) check

Answer 2

A vertical parabola has vertex (2,-8) and y intercept 4. What are its x intercepts?

The vertex form of the equation for the parabola is:

y + 8 = a(x - 2)²
y + 8 = ax² - 4ax + 4a
y = ax² - 4ax + 4a - 8

But the y intercept is 4.

4 = 4a - 8
12 = 4a
a = 3

y = ax² - 4ax + 4a - 8
y = 3x² - 12x + 12 - 8
y = 3x² - 12x + 4

Find the x intercepts.

3x² - 12x + 4 = 0
x = [12 ± √(144 - 4*3*4)] / (2*3)
x = [12 ± √(144 - 48)] / 6
x = [12 ± √96] / 6 = [12 ± 4√6] / 6
x = (6 ± 2√6] / 3

Answer 3

what book are you and what chapter