# Category: Graphing and Transformation

## How to sketch both cartesian and parametric graphs on the same diagram using GC

**Example **

Suppose we want to sketch the curve C and the line y=-2x+2 on the same graph

Note how the line y-2x+2 is entered.

Adjust the Tmax to 1% more so the graph displays correctly.

## Finding sequence of transformation

Usually the sequence of transformation can be found by the replacement method. However, it can be tough to find the replacement for complicated functions. For such cases, we can use the general linear transformation c f(bx+a)+d to find the sequence.

**cf(bx+a)+d **= Translate by a units in the negative X direction, then scale by a factor of 1/b parallel to the X-axis, then scale by a factor of c parallel to the Y-axis, then translate by d units in the positive Y direction.

**Example**

**Solution**

## Sketching Periodic functions with GC

Suppose we want to sketch the graph from -3 to 8.

- Enter by pressing “Math, Piecewise”.

2. Adjust the window setting to the required domain.

3. Trace the end point at -3 since it is not 1 complete cycle.

## How to check Sequence of Transformation?

## Advanced graph transformation

Advanced graph transformation can be done in stages, using the replacement method. Each time, we replace x or y.

## Challenging graphing technique question 1

This question was posted by a student at Singapore JC maths.

**Solution for Part ii**

## Application of graph transformation

## Graph transformation of oblique asymptote

Many students have difficulty with the graph transformation of oblique asymptote. Consider the oblique asymptote y = x-1 (red line)

i) y= 1/ f(x)

f(x) approaches infinity as x approaches infinity. 1 divided by infinity is 0.

Hence oblique asymptote y=x-1 becomes horizontal asymptote y= 0

**For y= 1/ f(x), any oblique asymptote y=ax+b in f(x) will become horizontal asymptote y= 0 **

ii) y= f ‘ (x)

The gradient of y= x-1 is 1. Hence oblique asymptote y=x-1 becomes horizontal asymptote y= 1

**For y= f ‘ (x), any oblique asymptote y=ax+b in f(x) will become horizontal asymptote y= a**

iii) y= f(2x+1)

Let original asymptote be f(x) = x-1

Therefore, f(2x+1) = (2x+1)-1 = 2x

Hence oblique asymptote y = x-1 is transformed to y=2x.

This method can be applied to any oblique asymptote.

## Reverse graph transformation

This question was posted by a student.

” What is the series of transformation from f(3-x/2) to f(x)?”

Here are 3 different methods to solve it.

**Method 1**

Let f(3-x/2) = g(x)

Let u=3-x/2

x= -2u+6

f(u)=g(-2u+6)

f(x)=g(-2x+6)

Therefore, the sequence of transformation is

Translate 6 units in the negative X direction.

Scale by a factor of 1/2 parallel to the X-axis.

Reflect about the Y-axis

**Method 2a**

Apply f(-2x) to f(3-0.5x).

f(-2x) = f[3-0.5(-2x)] = f(x+3)

ie. **Scale parallel to the X axis by a factor of 0.5. Then reflect about the Y axis**

Apply f(x-3) to f(x+3) to get f(x)

ie**. Translate 3 units in the positive X direction**

**Method 2b**

Apply f(x+6) to f(3-0.5x) to get f(-0.5x)

ie.** Translate 6 units in the negative X direction.**

Apply f(-2x) to f(-0.5x) = f(x)

ie.** Scale parallel to the X axis by a factor of 0.5. Then reflect about the Y axis.**

**Method 3**

f(3-x/2) is transforming f(x) by

Ａ：Translate 3 units in the negative X direction

B: Scale by a factor of 2 parallel to the X axis

C: Reflect about Y axis.

So to get back f(x), we reverse the transformation:

C’: Reflect about Y axis

B’: Scale by a factor of 1/2 parallel to the X-axis.

A’: Translate 3 units in the positive X direction

**Comments**

Method 1 and 2 are the forward approach. Method 3 is the reverse approach.

Method 1 is the preferred approach as it is easier and faster.

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