What is the y-value of the vertex of4x^2+8x-8

-12

Step-by-step explanation:

y = 4x^2 + 8x - 8

Put brackets around the first 2 terms and pull out the common factor

y = (4x^2 + 8x) - 8

y = 4(x^2 + 2x) - 8

Take 1/2 of the linear term (2x) and square it. Put the square inside the brackets.

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

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

You have added 4*1 inside the brackets. You must subtract that amount outside the brackets.

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

Notice that the trinomial inside the brackets is a perfect square. Combine the terms outside the brackets.

y = 4(x + 1)^2 - 12    

You have completed the square and you are finished.

The vertex is (-1, - 12)

The y value is - 12.

Just to confirm this, I have included the graph.

2x^2+5x+17

your welcome. : )