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: 23

Answer

The required parabola is shown below.

Work Step by Step

Use the steps shown below to determine the graph of the quadratic equation. Step 1: A quadratic function can be written as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ corresponding to which the graph is a parabola whose vertex is at $\left( h,k \right)$. Thus, the vertex of the parabola $f\left( x \right)=2{{\left( x-\left( -2 \right) \right)}^{2}}+\left( -1 \right)$ is $\left( h,k \right)=\left( -2,-1 \right)$. Step 2: The standard parabola is also symmetric with respect to the line $x=h$. Therefore, the provided parabola is symmetric to $x=-2$. Step 3: If $a>0$ , the parabola opens upward and if $a<0$ then the parabola opens downward. Also, if $\left| a \right|$ is small, the parabola opens more flatly than if $\left| a \right|$ is large. And, from the provided equation of the function, it is observed that graph opens upward as $a>0$. Step 4: So, we see that the above steps lead to the parabola that is open upwards, has a vertex at $\left( -2,-1 \right)$ and intercepts as $\left( \frac{\sqrt{2}}{2}-2,0 \right)\text{,}\left( -\frac{\sqrt{2}}{2}-2,0 \right)\text{ and }\left( 0,7 \right)$. Thus, the required parabola is shown above.
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