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: 29


The required parabola is shown below.

Work Step by Step

The quadratic function can be expressed as: $\begin{align} & f\left( x \right)={{x}^{2}}+3x-10 \\ & ={{x}^{2}}+3x+{{\left( \frac{3}{2} \right)}^{2}}-10-\frac{9}{4} \\ & ={{\left( x+\frac{3}{2} \right)}^{2}}-\frac{49}{4} \end{align}$ Use the steps shown below to determine the graph of the quadratic equation. Step 1: The 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 $f\left( x \right)={{\left( x-\left( -\frac{3}{2} \right) \right)}^{2}}+\left( -\frac{49}{4} \right)$ is $\left( h,k \right)=\left( -\frac{3}{2},-\frac{49}{4} \right)$. Step 2: The standard parabola is also symmetric with respect to the line $x=h$. So, the provided parabola is symmetric to $x=-\frac{3}{2}$. 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 $\left( -\frac{3}{2},-\frac{49}{4} \right)$ and intercepts as $\left( -5,0 \right)\text{,}\left( 2,0 \right)\text{ and }\left( 0,-10 \right)$. Thus, the required parabola is shown above.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.