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: 36


The required parabola is as shown below.

Work Step by Step

So, for the provided quadratic equation $f\left( x \right)=3{{x}^{2}}-2x-4$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets: $\begin{align} & a=3, \\ & b=-2, \\ & c=-4 \end{align}$ 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 downward 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: $\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{-2}{2\times 3} \\ & =\frac{1}{3} \end{align}$ And, the y-coordinate can be calculated as: $\begin{align} & y=3{{\left( \frac{1}{3} \right)}^{2}}-2\left( \frac{1}{3} \right)-4 \\ & =\frac{1}{3}-\frac{2}{3}-4 \\ & =-\frac{13}{3} \end{align}$ So, the vertex is $\left( \frac{1}{3},-\frac{13}{3} \right)$. Step 3: Above steps lead to the parabola that opens upwards and has a vertex at $\left( \frac{1}{3},-\frac{13}{3} \right)$ and it intersects x-axis at $\left( 1+\frac{\sqrt{13}}{3} \right)$ and $\left( 1-\frac{\sqrt{13}}{3} \right)$ , and y-axis at $-4$. Thus, the required parabola is as shown above.
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