## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 330: 38

#### Answer

The required parabola is as shown below. #### Work Step by Step

So, for the provided quadratic equation $f\left( x \right)=6-4x+{{x}^{2}}$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets: \begin{align} & a=1, \\ & b=-4, \\ & c=6 \end{align} And use the steps shown below to determine the graph of the quadratic equation. Step 1: Determine how the parabola opens: Note that a, the coefficient of ${{x}^{2}}$ , is -1. If $a>0$ , the parabola opens in the upward direction. Also, if $\left| a \right|$ is small, the parabola opens more flatly than if $\left| a \right|$ is large. Now, from the provided equation of the function, it is observed that the graph opens upwards as $a>0$. Step 2: Calculate the vertex: The x-coordinate can be calculated as: \begin{align} & x=-\frac{b}{2a} \\ & =-\frac{-4}{2\times 1} \\ & =2 \end{align} And the y-coordinate can be calculated as: \begin{align} & y\left( 2 \right)=6-4\left( 2 \right)+{{\left( 2 \right)}^{2}} \\ & =6-8+4 \\ & =2 \end{align} So, the vertex is $\left( 2,2 \right)$. Step 3: Above steps lead to the parabola that opens downwards and has a vertex at $\left( 2,2 \right)$ and y-axis at $6$. So, the required parabola is as shown above.

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