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Stretching and reflecting graphs

Stretching and reflecting graphs

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Summary

Stretching and reflecting graphs

​​In a nutshell

Stretching a graph is a transformation which makes a graph narrower or wider. This is achieved by multiplying the function by a constant which affects either the xx or yy coordinates. The position of the constant within the function affects how the function is stretched, and it is important to know how to draw and interpret these transformations.



Stretching graphs

When stretching graphs, the coordinates which are affected by the transformation depends on the position of the constant within the function. If the constant is inside the function, the xx coordinates of the function are affected. If the constant lies outside the function, the yy coordinates of the function are affected.​


The graph of y=af(x)y=af(x)  represents a stretch of the graph y=f(x)y=f(x) by a scale factor of aa in the yy​ axis. 

Take the graph of the function y=f(x)y=f(x):

y=2f(x)y=2f(x) stretches the graph by doubling all the yy coordinates. y=12f(x)y = \dfrac12f(x) stretches the graph by halving all the yy coordinates.



The graph of y=f(ax)y=f(ax) represents a stretch of the graph y=f(x)y=f(x) by a scale factor of 1a\dfrac1a in the horizontal axis.

Take the graph of y=f(x)y=f(x):​

y=f(2x)y=f(2x) stretches the graph by dividing all the xx coordinates by 22. y=(12x)y=\bigg(\dfrac12x\bigg)stretches the graph by multiplying all the xx coordinates by 22​.


Maths; Transformations; KS5 Year 12; Stretching and reflecting graphs


Example 1

f(x)=25x2f(x)=25-x^2. Find the:

a: xx-intercepts of the graph y=f(2x)y=f(2x)

b: maximum height of the curve y=2f(x)y=2f(x)


a: f(x)=25x2f(x)=25-x^2

Factorise: 

f(x)=(5x)(5+x)f(x)=(5-x)(5+x)


xx-intercept is when f(x)=0f(x)=0:

(5x)(5+x)=0(5-x)(5+x)=0

x=5, x=5x=5,\ x=-5

​​​

When y=f(2x)y=f(2x), the xx coordinates of the curve are multiplied by 12\dfrac12, so:


The xx-intercepts of y=f(2x)y=f(2x) are (2.5,0) and (2.5,0)\underline{(2.5,0)\ and\ (-2.5,0)}.

This can also be seen by drawing the function:

Maths; Transformations; KS5 Year 12; Stretching and reflecting graphs

b: Maximum height of the curve is at the turning point. To find the turning point you must first differentiate the function​: 

ddx(25x2)=2x\dfrac{d}{dx} (25-x^2) = -2x

​​

At the turning point, dydx=0\dfrac{dy}{dx}=0:

2x=0, x=0-2x=0,\ x=0​​


​When x=0x=0:

y=25x2y=2502y=25y=25-x^2\\y=25-0^2\\y=25​​


When y=2f(x)y=2f(x), the yy coordinates of the curve are multiplied by 22, so:


The maximum height of the curve y=2f(x)y=2f(x) is 50\underline{50}.



Reflecting graphs

Reflecting a graph is a form of stretching which mirrors the graph in one of the axes. To reflect a graph you multiply the function by 1-1. The axis that the function is reflected in is based on whether the constant is in the inside or on the outside of the function.


The function y=f(x)y=-f(x) is a reflection of the graph y=f(x)y=f(x) in the xx-axis. It is the same as multiplying all of the yy values by 1-1.​


Maths; Transformations; KS5 Year 12; Stretching and reflecting graphs


The function y=f(x)y=f(-x) is a reflection of the graph y=f(x)y=f(x) in the yy-axis. It is the same as multiplying all of the xx values by 1-1.​


Maths; Transformations; KS5 Year 12; Stretching and reflecting graphs



Example 2

A quadratic function y=f(x)y=f(x) intercepts the xx-axis at x=0x=0 and x=7x=7, and has a minimum point at y=13y=-13. What are the xx-intercepts and the values of yy at the minimum/maximum point of the curves:

a: y=f(x)y=-f(x)

b: y=f(x)y=f(-x)


a: When y=f(x)y=-f(x), the function is reflected in the xx-axis. The xx values remain the same but the yy values are multiplied by 1-1, therefore: 


The xx-intercepts are (0,0) and (7,0)\underline{(0,0)\ and\ (7,0)} and the maximum point is at y=13\underline{y=13}.


b: When y=f(x)y=f(-x), the function is reflected in the yy-axis. The yy values remain the same but the xx values are multiplied by 1-1, therefore:


The xx-intercepts are (0,0) and (7,0)\underline{(0,0)\ and\ (-7,0)} and the minimum point is at y=13\underline{y=-13}.




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