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 $x$ or $y$ 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 $x$ coordinates of the function are affected. If the constant lies outside the function, the $y$ coordinates of the function are affected.

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

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

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

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

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

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

$y=f(2x)$ stretches the graph by dividing all the $x$ coordinates by $2$ . $y=\bigg(\dfrac12x\bigg)$ stretches the graph by multiplying all the $x$ coordinates by $2$ .

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

Example 1

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

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

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

a: $f(x)=25-x^2$

Factorise:

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

$x$ -intercept is when $f(x)=0$ :

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

$x=5,\ x=-5$

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

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

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:

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

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

$-2x=0,\ x=0$ 

When $x=0$ :

$y=25-x^2\\y=25-0^2\\y=25$

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

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

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

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

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

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

Example 2

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

a: $y=-f(x)$

b: $y=f(-x)$

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

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

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

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