## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 330: 29

#### Answer

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.

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