## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 8 - Section 8.5 - Quadratic Functions and Their Graphs - Exercise Set: 70

#### Answer

$f(x)=5(x-1)^{2}+6$

#### Work Step by Step

The vertex-form of a quadratic equation is $y=a(x-h)^{2}+k$, with a vertex at the opposite of $h$ and $k$, or $(h, k)$. The original equation in this problem is $f(x)=5x^{2}$; written in vertex-form it is $f(x)=5(x-0)^{2}+0$ and has a vertex of $(0, 0)$. To translate this parabola so that the vertex is at $(1, 6)$ we need to rewrite the equation as $f(x)=5(x-1)^{2}+6$.

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