Chapter 1, Problem 14PS

a.

To determine

Sketch the graph of each function.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}f(1)=a{(}^1+b(1)+2\\ 0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $-2x^2+2$ .

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

  $\begin{array}{l}f\left(x+1\right)=-2{\left(x+1\right)}^2+2\\ =-2\left(x^2+1+2x\right)+2\\ =-2x^2-\cancel{2}-4x+\cancel{2}\\ =-2x^2+4x\end{array}$

Sketch the above graph using maple software as done below,

  

b.

To determine

Sketch the graph of each function.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $-2x^2+2$ .

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

  $\begin{array}{l}f\left(x\right)+1=-2+2+1\\ =-2x^2+3\end{array}$

The graph will be shifted $1$ unit upwards.

Sketch the above graph using maple software as done below,

  

c.

To determine

Sketch the graph of the function.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $-2x^2+2$ .

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

  $\begin{array}{l}2f\left(x\right)=2(-2x^2+2)\\ =-4x^2+4\end{array}$

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

  

d.

To determine

Sketch the graph of each function.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $\boxed{-2x^2+2}$ .

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

  $\begin{array}{l}f\left(-x\right)=-2{(}^{-}+2\\ =-2x^2+2\end{array}$

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

  

e.

To determine

Plot the graph.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $-2x^2+2$ .

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

  $\begin{array}{l}-f\left(x\right)=-\left(-2x^2+2\right)\\ =2x^2+2\end{array}$

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

  

f.

To determine

Plot the graph.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $\boxed{-2x^2+2}$ .

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

  

f.

To determine

Plot the graph.

Answer to Problem 14PS

  

Explanation of Solution

Calculation:

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

  $\begin{array}{l}h\left(x\right)=f\left(x\right)+c\left(upwards\right)\\ h\left(x\right)=f\left(x\right)-c\left(downwards\right)\end{array}$

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

  $\begin{array}{l}h\left(x\right)=f\left(x+c\right)\left(shift,left\right)\\ h\left(x\right)=f\left(x-c\right)\left(shift,right\right)\end{array}$

Consider the following graph,

  

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

  $f\left(x\right)=a{x}^2+bx+c$

Observe the above function and following data is obtained,

  $f\left(0\right)=2,f\left(1\right)=0,f\left(-1\right)=0$

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

  $\begin{array}{l}f\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 2=c\end{array}$

Substitute this value in the general function $f\left(x\right)=a{x}^2+bx+c$ , hence it becomes,

  $f\left(x\right)=a{x}^2+bx+c$

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

  $\begin{array}{l}$ f(1)=a{(1)}^2+b(1)+2$0=a+b+\mathrm{2......}(1)\\ a+b=-2\end{array}$

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

  $f\left(x\right)=-2x^2+2$

Hence, the function of the above graph is $\boxed{-2x^2+2}$ .

Consider the function $f\left|\left(x\right)\right|=-2{\left|x\right|}^2+2$ , 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,