Slippery Slopes

Thank you to Sophie, Mimi, Hannah and Emma from Sandbach High School who submitted the answers to this problem in the diagrams below:






Thank you also to Melissa who has given explanations for how she arrived at her answers:

First part:

The graphs are all transformations of $y = f(x)$ so $y = 3 f(x)$ and $y = 3f(x) + 8$ can be easily spotted as the lines with steeper gradients, with $y = 3f(x) + 8$ being higher up on the graph. Similarly, of $y = f(x)$ and $y = f(x) - 8$, $y = f(x) - 8$ is the line which is seen to be lower down on the graph. So the yellow, black, blue and red lines show $y = 3f(x) + 8$, $y = 3f(x)$, $y = f(x)$ and $y = f(x) - 8$ respectively.

A and B will have gradients three times as large as the gradients of C and D because for each increase by one in the x value, 3f(x) increases by three times as much as f(x) does.

This means that when the gradient of D is $\frac{1}{4}$, the gradient of C is also $\frac{1}{4}$ and the gradients of A and B are $3\times\frac{1}{4} = \frac{3}{4}$.

Second part:

The yellow line has been reflected in the x axis compared with the others because it starts off decreasing whereas the others begin increasing. Therefore the yellow line is $y = -3f(x) + 10$.
The black line has the same gradient as the yellow line, which is a steeper gradient than the others so the black shows $y = 3f(x - 25)$.
The red line is further left than the blue line so the red line shows $y = f(x + 20)-8$ and the blue line shows $y = f(x)$.

The points E, F, G and H are the same point translated from the original curve, which can easily be seen by looking at how far horizontally and vertically the points E, F and G have been translated from H.

The gradient at E is $-\frac{1}{4}$ so the gradient at G is also $-\frac{1}{4}$ because $y = f(x + 20)$ has the same shape as $y = f(x)$, it is translated but not stretched. The gradient at H is $-\frac{3}{4}$ because the graph $y = 3f(x - 25)$ has been stretched along the y axis by a scale factor of 3 from $y = f(x)$ so the gradient is three times that of G. The gradient at F is $\frac{3}{4}$ because the gradient is the same as at H but is positive.