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

User Jrgm
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1 Answer

4 votes

Answer:

-12

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.

What is the Y-value of the vertex of4x^2+8x-8-example-1
User Juv
by
5.8k points