What is the y value of the vertex of 4x^2+8x-8

To find the vertex, [tex](h,k)[/tex], or a quadratic of the form [tex]ax^2+bx+c[/tex], we are going to use the vertex formula: [tex]h= \frac{-b}{2a} [/tex], and then, we are going to evaluate the function at [tex]h[/tex] to find [tex]k[/tex]:

We can infer four our quadratic 4x^2+8x-8 that [tex]a=4[/tex] and [tex]b=8[/tex], so lets replace those values in our formula to find [tex]h[/tex]:
[tex]h= \frac{-b}{2a} [/tex]
[tex]h= \frac{-8}{(2)(4)} [/tex]
[tex]h= \frac{-8}{8} [/tex]
[tex]h=-1[/tex]

To find [tex]k[/tex] we are going to evaluate the quadratic at [tex]h=-1[/tex]. In other words, we are going to replace [tex]x[/tex] with -1:
[tex]4x^2+8x-8[/tex]
[tex]k=4(-1)^2+8(-1)-8[/tex]
[tex]k=4-8-8[/tex]
[tex]k=-12[/tex]

Our vertex is (-1,-12). We can conclude that the y value of the vertex of 4x^2+8x-8 is -12.

calculista calculista · Answer 2 · 2019-02-03T23:45:39+01:00

Answer:

The y-value of the vertex is [tex]-12[/tex]

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

[tex]f(x)=a(x-h)^{2}+k[/tex]

where

(h,k) is the vertex of the parabola

In this problem we have

[tex]f(x)=4x^{2}+8x-8[/tex] -----> this a vertical parabola open upward

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

[tex]f(x)+8=4x^{2}+8x[/tex]

Factor the leading coefficient

[tex]f(x)+8=4(x^{2}+2x)[/tex]

Complete the square. Remember to balance the equation by adding the same constants to each side

[tex]f(x)+8+4=4(x^{2}+2x+1)[/tex]

[tex]f(x)+12=4(x^{2}+2x+1)[/tex]

Rewrite as perfect squares

[tex]f(x)+12=4(x+1)^{2}[/tex]

[tex]f(x)=4(x+1)^{2}-12[/tex]

The vertex is the point [tex](-1,-12)[/tex]

The y-value of the vertex is [tex]-12[/tex]