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

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To start with, you need to find the vertex, you need to find the axis of symmetry: x=(-b/2a).  b=8.  a=4.  x=(-8/2(4)).  x=-1.  Plugin -1 into the original equation to find y.  y=-12.
User Barin
by
7.8k points
1 vote

Answer:

The y-value of the vertex is
-12

Explanation:

we know that

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


f(x)=a(x-h)^(2)+k

where

(h,k) is the vertex of the parabola

In this problem we have


f(x)=4x^(2)+8x-8 -----> 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


f(x)+8=4x^(2)+8x

Factor the leading coefficient


f(x)+8=4(x^(2)+2x)

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


f(x)+8+4=4(x^(2)+2x+1)


f(x)+12=4(x^(2)+2x+1)

Rewrite as perfect squares


f(x)+12=4(x+1)^(2)


f(x)=4(x+1)^(2)-12

The vertex is the point
(-1,-12)

The y-value of the vertex is
-12

User Sakana
by
7.8k points

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