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: 34

Answer

The required parabola is as shown below.

Work Step by Step

So for the provided quadratic equation $f\left( x \right)={{x}^{2}}+4x-1$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets: $\begin{align} & a=1, \\ & b=4, \\ & c=-1 \end{align}$ 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 upwards and if $a<0$ then the parabola opens downwards. 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: $\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{4}{2\times 1} \\ & =-2 \end{align}$ Or, the y-coordinate can be calculated as: $\begin{align} & y={{\left( -2 \right)}^{2}}+4\left( -2 \right)-1 \\ & =4-8-1 \\ & =-5 \end{align}$ Therefore, the vertex is $\left( -2,-5 \right)$. Step 3: So, the above steps lead to the parabola that opens upwards and has a vertex at $\left( -2,-5 \right)$ and it intersects x-axis at $\left( -2+\sqrt{5} \right)$ and $\left( -2-\sqrt{5} \right)$ , and y-axis at -1. Thus, the required parabola is as shown above.
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