Question

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

Answer 1

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

Explanation:

y = `4x^2` + 4x - 8
y = 4(`x^2` + x) - 8
y = 4(`x^2` + x + _ - _ ) - 8
___ = `(b/2)^2`
___ = `(1/2)^2`

___ = `1/4`

y = 4(`x^2` + x + `1/4` - `1/4`) - 8

y = 4(`x^2` + x + `1/4`) - 1 - 8

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

In the form y = a`(x - p)^2` + q, the vertex is found at (p,q). So, in y = 4`(x +`1/2`)^2` - 9, the vertex is found at (-`1/2`, -9).

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

y = `4(0)^2` + 4(0) - 8
y = -8

In other words, in quadratic functions of form y = `ax^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 = `4x^2` + 4x - 8
0 = `4x^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 = `4x^2` + 4x - 8is at (`-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.