Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.1 Quadratic Functions and Parabolas - 4.1 Exercises - Page 303: 45

Answer

The vertex must be on the $x$-axis.

Work Step by Step

This question is asking us to design a parabola with one horizontal intercept. Start with the vertex form of a parabola and set $h= 2$ and $k= 0$. $$\begin{aligned} f(x) &= a(x-h)^2+ k=a(x-2)^2+0=a(x-2)^2. \end{aligned}$$ Since the parabola opens upward, the value of the constant $ a$ must be positive. For $a= 2$, we have. $$\begin{aligned} y &= 2(x-2)^2. \end{aligned}$$ The vertex must be on the $x$-axis.
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