4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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The word ‘transformation’ means modification. Transformation of the graph of a function is a process by which we modify or change the original graph and make a new graph.

Exponential Functions

The exponential function is a type of mathematical function which is used in real-world contexts. It helps to find out the exponential decay model or exponential growth model, in mathematical models. In this topic, we will understand descriptive rules, c…

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Question

Chapter 1, Problem 14PS

To determine

Sketch the graph of each function.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 1

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 2

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a(1)2+b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Now consider the function $f(x+1)$ , obtain the same by replacing $x$ by $(x+1)$ in the above function hence,

$f(x+1)=−2(x+1)2+2=−2(x2+1+2x)+2=−2x2−2−4x+2=−2x2+4x$

Sketch the above graph using maple software as done below,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 3

To determine

Sketch the graph of each function.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 4

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 5

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Now consider the function $f(x)+1$ , obtain the same by adding just $1$ in the above function hence,

$f(x)+1=−2+2+1=−2x2+3$

The graph will be shifted $1$ unit upwards.

Sketch the above graph using maple software as done below,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 6

To determine

Sketch the graph of the function.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 7

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 8

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Consider the function $2f(x)$ , this function is obtained magnifying the above graph by $2$ ,

$2f(x)=2(−2x2+2)=−4x2+4$

The graph of $2f(x)$ will look like below,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 9

To determine

Sketch the graph of each function.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 10

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 11

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Consider the function $f(−x)$ , this function is obtained by replacing $x$ by- $x$ the above graph,

$f(−x)=−2(−x)2+2=−2x2+2$

The graph of $f(−x)$ will exactly be same as $f(x)$ hence it will look like,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 12

To determine

Plot the graph.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 13

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 14

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Consider the function $−f(x)$ , this function is obtained by inverting the above original graph,

$−f(x)=−(−2x2+2)=2x2+2$

The graph of $−f(x)$ will exactly be same as $−f(x)$ hence it will look like,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 15

To determine

Plot the graph.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 16

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 17

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Consider the function $|f(x)|=|−2x2+2|$ , all negative values will get reflected about x-axis and hence the function will look like below,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 18

To determine

Plot the graph.

Expert Solution

Answer to Problem 14PS

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 19

Explanation of Solution

Calculation:

The general rule of the function shifting vertically is as below,

$h(x)=f(x)+c(upwards)h(x)=f(x)−c(downwards)$

The general rule of the function shifting horizontally upwards is as below,

$h(x)=f(x+c)(shift,left)h(x)=f(x−c)(shift,right)$

Consider the following graph,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 20

The above function looks like qudratic one end hence assume it as,

$f(x)=ax2+bx+c$

Observe the above function and following data is obtained,

$f(0)=2,f(1)=0,f(−1)=0$

Substitute each of the baove values on the general function as done below,

$f(0)=a(0)2+b(0)+c2=c$

Substitute this value in the general function $f(x)=ax2+bx+c$ , hence it becomes,

$f(x)=ax2+bx+c$

Substitute $f(1)=0$ in the above equation,

$f(1)=a (1) 2 +b(1)+20=a+b+2......(1)a+b=−2$

Solve (1) and (2), so that the values of constants are $a=−2,b=0$ hence the quadratic function becomes,

$f(x)=−2x2+2$

Hence, the function of the above graph is $−2x2+2$ .

Consider the function $f|(x)|=−2|x|2+2$ , the absolute value of the function does not matter anything here as the value of $x$ is already squared, hence the function will not have any change and it will look like below,

EBK PRECALCULUS W/LIMITS, Chapter 1, Problem 14PS , additional homework tip 21

Chapter 1, Problem 13PS

Chapter 1, Problem 15PS

Chapter 1 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E

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