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Contact [7]
3 years ago
10

Write f(x)=8x^2-4x+11 in vertex form

Mathematics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2}

or

f(x)=8(x-0.25)^{2}+10.5

Step-by-step explanation:

we have

f(x)=8x^{2}-4x+11

This is a vertical parabola open upward (because the leading coefficient is positive)

The vertex is a minimum

Convert to vertex form

Factor the leading coefficient

f(x)=8(x^{2}-\frac{1}{2}x)+11

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

f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+11-\frac{1}{2}

f(x)=8(x^{2}-\frac{1}{2}x+\frac{1}{16})+\frac{21}{2}

Rewrite as perfect squares

f(x)=8(x-\frac{1}{4})^{2}+\frac{21}{2} ----> equation in vertex form

or

f(x)=8(x-0.25)^{2}+10.5

The vertex is the point  (0.25,10.5)

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