2 years ago

Alex bought soccer shoes on sale

Mathematics

2 answers:

Radda [10]2 years ago

6 0

He spent $34.50 in all

mr Goodwill [35]2 years ago

4 0

He spent 34.50 dollars in total

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

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Answer:

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

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

Step-by-step explanation:

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

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

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

The vertex is the point (0.25,10.5)

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