Chapter 3.5, Problem 53HP

To determine

To find: Horizontal translation of $f\left(x\right)=mx+b$ to $g\left(x\right)=mx+b+k$

Answer to Problem 53HP

  $f\left(x\right)$ is horizontally shifted by $\frac{k}{m}$ units towards left to obtain the same line $g\left(x\right)=mx+b+k$ .

Explanation of Solution

Given information:

  $f\left(x\right)=mx+b$

  $g\left(x\right)=mx+b+k$ is obtained by vertical translation of $k$ units upwards along Y-axis.

Concept:

A Translation is a movement of the graph either horizontally parallel to x-axis or vertically parallel to y-axis.

The graph of $y=f\left(x\right)$ where $f\left(x\right)=x$ is same as the graph of $y=x$ . Writing the equations as functions in the form of $f\left(x\right)$ is useful when applying translations and reflections to graphs.

If $g\left(x\right)=f\left(x\right)+k$ it is known as vertical shift.

If $k>0$ it moves upwards.

If $k<0$ it moves downwards.

If $g\left(x\right)=f\left(x+l\right)$ it is known as horizontal shift.

If $l<0$ moves it right.

If $l>0$ moves it left.

If $g\left(x\right)=-f\left(x\right)$ it is known as Vertical reflection it means reflects about X-axis

If $g\left(x\right)=f\left(-x\right)$ it is known as Horizontal reflection it means reflects about Y-axis

If $g\left(x\right)=af\left(x\right)$

If $a>1$ it is known as Vertical stretch.

If $0<a<1$ it is known as Vertical compression.

If $g\left(x\right)=f\left(bx\right)$

If $b>1$ it is known as Horizontal compression.

If $0<b<1$ it is known as Horizontal Stretch.

Calculation:

We know that

If $g\left(x\right)=f\left(x+l\right)$ it is known as horizontal shift.

If $l<0$ moves it right.

If $l>0$ moves it left.

So, to translate the function $f\left(x\right)=mx+b$ to $g\left(x\right)=mx+b+k$ we should write $g\left(x\right)$ as of the form $f\left(x+l\right)$ .

  $g\left(x\right)=f\left(x+l\right)$

  $\begin{array}{l}\Rightarrow mx+b+k=m\left(x+l\right)+b\\ \Rightarrow mx+b+k=mx+ml+b\\ \Rightarrow \left(mx+b\right)+k=\left(mx+b\right)+ml\\ \Rightarrow k=ml\\ \Rightarrow l=\frac{k}{m}\end{array}$

Since given $k$ is positive, i.e. $k>0$ which also implies that $\frac{k}{m}>0$

Therefore, $l>0$

So the function $f\left(x\right)$ has to be translated using horizontal translation (shift) of graph of the function $f\left(x\right)$ by $l$ units i.e. $\frac{k}{m}$ units towards left to obtain the same line $g\left(x\right)=mx+b+k$

  $\Rightarrow g\left(x\right)=f\left(x+\frac{k}{m}\right)$ which is horizontal shift of $f\left(x\right)$ by $\frac{k}{m}$ units towards left.