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: 6

Answer

The equation of the given graph is\[f\left( x \right)={{x}^{2}}+2x+1\]

Work Step by Step

A quadratic function can be expressed as $f\left( x \right)=a{{x}^{2}}+bx+c$ corresponding to which the graph is a parabola whose vertex is the point $\left( h,k \right)=\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$. If $a>0$ , the parabola opens upward and if $a<0$ then the parabola opens downward. Now, it can be observed that the given parabola opens upward so $a>0$. So, the vertex of the parabola in the graph is on $\left( -1,0 \right)$. Hence, in comparison with the standard equation, the value of $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$ is: $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)=\left( -1,0 \right)$ This implies, $\begin{align} & \frac{-b}{2a}=-1 \\ & b=2a \end{align}$ and, $f\left( -1 \right)=0$ Putting in the values in the standard form to we get: $\begin{align} & f\left( -1 \right)=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\ & 0=a-b+c \\ & b=a+c \end{align}$ Putting in the values in the standard form we get: $f\left( x \right)=a{{x}^{2}}+\left( a+c \right)x+c$ Where, $a>0$, the parabola opens upwards. The coefficient of $x$ is the sum of the coefficient of ${{x}^{2}}$ and has the same sign of x. This equation is similar to $f\left( x \right)={{x}^{2}}+2x+1$. Therefore, the required equation of the graph is $f\left( x \right)={{x}^{2}}+2x+1$.
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