Q. 92 - Precalculus Enhanced with Graphing Utilities

Question

Suppose $\left(1,3\right)$ is a point on the graph of $y=f\left(x\right)$.

(a) What point is on the graph of$y=f(x+3)-5$?

(b) What point is on the graph of $y=-2f\left(x-2\right)+1$?

(c) What point is on the graph of $y=\left(2x+3\right)$?

Step-by-Step Solution

Answer

Part (a). The point $\left(-2,-2\right)$ is on the graph $y=f(x+3)-5$.

Part (b). The point $\left(3,-5\right)$ is on the graph $y=-2f(x-2)+1$.

Part (c). The point $\left(-1,3\right)$ is on the graph $y=f(2x+3)$

Part (a) Step 1. Finding the point on y = f ( x + 3 ) - 5

It is given that $(1,3)$ is a point on the graph $y=f\left(x\right)$.

So, substituting $x+3=1$ in the function

$y=f(x+3)-5$:

$$\begin{gathered} x+3=1 \\ x=1-3 \\ x=-2 \end{gathered}$$

Now, we have:

$$\begin{gathered} y=f\left(1\right)-5 \\ y=3-5 \left[f\left(1\right)=3\right] \\ y=-2 \end{gathered}$$

Hence, $\left(-2,-2\right)$ is on the graph $y=f(x+3)-5$

Part (b) Step 1. Finding the point on y = - 2 f ( x - 2 ) + 1

It is given that $\left(1,3\right)$ is a point on the graph $y=f\left(x\right)$.

So, substituting $x-2=1$ in the function

$y=-2(x-2)+1$:

$$\begin{gathered} x-2=1 \\ x=1+2 \\ x=3 \end{gathered}$$

Now, we have:

$$\begin{gathered} y=-2f\left(1\right)+1 \\ y=-2\left(3\right)+1 \\ y=-6+1 \\ y=-5 \end{gathered}$$

Hence, $\left(3,-5\right)$ is on the graph $y=-2f(x-2)+1$

Part (c) Step 1. Finding the point on y = f 2 x + 3

It is given that $\left(1,3\right)$ is a point on the function $y=f\left(x\right)$.

So, substituting $2x+3=1$ in the function $y=f(2x+3)$

$$\begin{gathered} 2x+3=1 \\ 2x=1-3 \\ 2x=-2 \\ x=-1 \end{gathered}$$

Now, we have:

$$\begin{gathered} y=f\left(1\right) \\ y=3 \left[f\left(1\right)=3\right] \end{gathered}$$

Hence, $\left(-1,3\right)$ is on the graph $y=f(2x+3)$