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: 19

Answer

The graph is shown below. Range $[2,\infty)$.

Work Step by Step

The given function is $f(x)=(x-1)^2+2$ The standard form of the parabola is $f(x)=a(x-h)^2+k$ Compare with the original given function. $a=1,h=1$ and $k=2$ Step 1:- Parabola opens. $a>0$, The parabola opens upward. Step 2:- Vertex. The value of $h=1$ and $k=2$. The vertex is $(h,k)=(1,2)$. Step 3:- $x-$intercepts. Replace $f(x)$ with $0$. $\Rightarrow 0=(x-1)^2+2$ Subtract $2$ from both sides. $\Rightarrow 0-2=(x-1)^2+2-2$ Simplify. $\Rightarrow -2=(x-1)^2$ Apply the square root property. $\Rightarrow x-1=\sqrt{-2}$ or $x-1=-\sqrt{-2}$ Hence, the equation has imaginary solutions, there are no $x-$intercepts. Step 4:- $y-$intercept. Replace $x$ with $0$. $\Rightarrow f(0)=(0-1)^2+2$ Simplify. $\Rightarrow f(0)=1+2$ Simplify. $\Rightarrow f(0)=3$ Hence, the $y-$intercept is $3$. The parabola passes through $(0,3)$. Step 5:- Graph. Use the points vertex and $y-$intercept to draw a parabola. The axis of symmetry is $x=h=1$. $A=(1,2)$ $B=(0,3)$. From the graph the range of the function is $[2,\infty)$.
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