Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.3 - Quadratic Functions and Their Graphs - Exercise Set - Page 626: 32

Answer

The graph is shown below. Range $(-\infty,9]$.

Work Step by Step

The given function is a quadratic function: $f(x)=5-4x-x^2$ Rewrite as $f(x)=-x^2-4x+5$ The standard form of the quadratic function is $f(x)=ax^2+bx+c$ Compare both equations $a=-1,b=-4$ and $c=5$. Step 1:- Parabola opens. $a<0$, the parabola open downward. Step 2:- Vertex. $x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{(-4)}{2(-1)}=-2$. Substitute the value of $x$ into the function. $\Rightarrow f(-2)=-(-2)^2-4(-2)+5$ Simplify. $\Rightarrow f(-2)=-4+8+5$ $\Rightarrow f(-2)=9$ The vertex is $(-2,9)$. Step 3:- $x-$intercepts. Replace $f(x)$ with $0$ into the given function. $\Rightarrow 0=-x^2-4x+5$ Use quadratic formula, we have $a=-1,b=-4$ and $c=5$. $\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Substitute all values. $\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^2-4(-1)(5)}}{2(-1)}$ Simplify. $\Rightarrow x=\frac{4\pm\sqrt{16+20}}{-2}$ $\Rightarrow x=\frac{4\pm\sqrt{36}}{-2}$ $\Rightarrow x=\frac{4\pm6}{-2}$ Separate the fractions. $\Rightarrow x=\frac{4+6}{-2}$ or $x= \frac{4-6}{-2}$ Simplify. $\Rightarrow x=\frac{10}{-2}$ or $ x=\frac{-2}{-2}$ $\Rightarrow x=-5$ or $ x=1$ The $x-$intercepts are $-5$ and $1$. The parabola passes through $(-5,0)$ and $(1,0)$. Step 4:- $y-$intercept. Replace $x$ with $0$ in the given function. $\Rightarrow f(0)=-(0)^2-4(0)+5$ Simplify. $\Rightarrow f(0)=5$ The $y-$intercept is $5$. The parabola passes through $(0,5)$. Step 5:- Graph. Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola. The axis of symmetry is $x=-2$. $A=(-2,9)$ $B=(-5,0)$ $C=(1,0)$ $D=(0,5)$. From the graph the range of the function is $(-\infty,9]$.
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