Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 330: 56

Answer

The standard form is \[f\left( x \right)=3{{\left( x-9 \right)}^{2}}\]

Work Step by Step

We know that a quadratic function can be expressed in the standard form as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ corresponding to which the graph is a parabola whose vertex is the point $\left( h,k \right)$. The parabola attains its minimum value of 0 so it must be opening upwards. We have $a>0$ , therefore, $f\left( h \right)=k$ is the minimum. Therefore, $k=0$. Since the parabola attains its maximum at 9, thus $h=9$ The parabola is upwards so the shape of the parabola is $f\left( x \right)=3{{x}^{2}}$. Now, put the obtained results in the standard form to get: $\begin{align} & f\left( x \right)=3{{\left( x-h \right)}^{2}}+k \\ & =3{{\left( x-9 \right)}^{2}}+0 \\ & =3{{\left( x-9 \right)}^{2}} \end{align}$ Hence, the parabola is expressed in standard form as $f\left( x \right)=3{{\left( x-9 \right)}^{2}}$.
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