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: 29

Answer

The graph is shown below. Range $\left[-\frac{49}{4},\infty\right)$.

Work Step by Step

The given function is a quadratic function: $f(x)=x^2+3x-10$ The standard form of the quadratic function is $f(x)=ax^2+bx+c$ Compare both equations $a=1,b=3$ and $c=-10$. Step 1:- Parabola opens. $a>0$, the parabola open upward. Step 2:- Vertex. $x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{3}{2(1)}=-\frac{3}{2}$. Substitute the value of $x$ into the function. $\Rightarrow f(-\frac{3}{2})=(-\frac{3}{2})^2+3(-\frac{3}{2})-10$ Simplify. $\Rightarrow f(-\frac{3}{2})=\frac{9}{4}-\frac{9}{2}-10$ The LCD is $4$. Multiply the numerator and the denominator to form LCD at the denominator. $\Rightarrow f(-\frac{3}{2})=\frac{9}{4}-\frac{18}{4}-\frac{40}{4}$ Add all numerators. $\Rightarrow f(-\frac{3}{2})=\frac{9-18-40}{4}$ Simplify. $\Rightarrow f(-\frac{3}{2})=-\frac{49}{4}$ The vertex is $(-\frac{3}{2},-\frac{49}{4})$. Step 3:- $x-$intercepts. Replace $f(x)$ with $0$ into the given function. $\Rightarrow 0=x^2+3x-10$ Use the quadratic formula, we have $a=1,b=3$ and $c=-10$. $\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Substitute all values. $\Rightarrow x=\frac{-(3)\pm\sqrt{(3)^2-4(1)(-10)}}{2(1)}$ Simplify. $\Rightarrow x=\frac{-3\pm\sqrt{9+40}}{2}$ $\Rightarrow x=\frac{-3\pm\sqrt{49}}{2}$ $\Rightarrow x=\frac{-3\pm7}{2}$ Separate the fractions. $\Rightarrow x=\frac{-3+7}{2}$ or $ \frac{-3-7}{2}$ Simplify. $\Rightarrow x=\frac{4}{2}$ or $ x=\frac{-10}{2}$ $\Rightarrow x=2$ or $ x=-5$ The $x-$intercepts are $2$ and $-5$. The parabola passes through $(2,0)$ and $(-5,0)$. Step 4:- $y-$intercept. Replace $x$ with $0$ in the given function. $\Rightarrow f(0)=(0)^2+3(0)-10$ Simplify. $\Rightarrow f(0)=-10$ The $y-$intercept is $-10$. The parabola passes through $(0,-10)$. Step 5:- Graph. Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola. The axis of symmetry is $x=-\frac{3}{2}$. $A=(-\frac{3}{2},-\frac{49}{4})$ $B=(2,0)$ $C=(-5,0)$ $D=(0,-10)$. From the graph the range of the function is $\left[-\frac{49}{4},\infty\right)$.
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