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

1 Answer
Dec 30, 2015

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.