# How do you find the vertex, x and y intercepts for y=4x^2+4x-8?

Dec 30, 2015

First change to vertex form by completing the square in order to easily find the vertex

#### Explanation:

y = $4 {x}^{2}$ + 4x - 8
y = 4(${x}^{2}$ + x) - 8
y = 4(${x}^{2}$ + x + _ - _ ) - 8
= ${\left(\frac{b}{2}\right)}^{2}$
= ${\left(\frac{1}{2}\right)}^{2}$

___ = $\frac{1}{4}$

y = 4(${x}^{2}$ + x + $\frac{1}{4}$ - $\frac{1}{4}$) - 8

y = 4(${x}^{2}$ + x + $\frac{1}{4}$) - 1 - 8

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

In the form y = a${\left(x - p\right)}^{2}$ + q, the vertex is found at (p,q). So, in y = 4(x + 1/2)^2 - 9, the vertex is found at (-$\frac{1}{2}$, -9).

The y intercept can be found by plugging in 0 in x's place:

y = $4 {\left(0\right)}^{2}$ + 4(0) - 8
y = -8

In other words, in quadratic functions of form y = $a {x}^{2}$ + bx + c, the y intercept is always (0, c).

As for the x intercept, you find it by plugging in 0 for y and solving the resulting quadratic equation using the quadratic formula or by factoring when possible:

0 = $4 {x}^{2}$ + 4x - 8
0 = $4 {x}^{2}$ + 8x - 4x - 8
0 = 4x(x+ 2) - 4(x + 2)
0 = (4x -4)(x + 2)
x = 1 and -2.

The x intercepts are at (1, 0) and (-2,0)

To summarize, the vertex of y = $4 {x}^{2}$ + 4x - 8is at ($- \frac{1}{2}$, -9), the y intercept is at (0, -8) and the x intercepts are at (1,0) and (-2,0).

I hope you understand now.