Reconstruct signs from two absolute-value sketches

y=\frac1{x-3}

Total marks: 7

Solution

(a)i

Describing transformations

Describe a sequence of exactly three transformations that will transform the curve $y=\dfrac1{x-3}$ onto the curve $y=\dfrac{2x+a}{3-x}$ , where $a$ is a positive constant.

Explore this idea

Hint

Write $\frac{2x+a}{3-x}$ in the form $\text{constant}+\frac{\text{constant}}{3-x}$ by splitting the fraction.
Compare it with $\frac1{x-3}$ and name each operation: reflect, stretch, translate.

Answer · (a)i

a+6

Explore the idea

Reflect, stretch, shift

Drawn with a = 2, so the stretch factor is 6 + a = 8.

Start: y = 1/(x − 3)

The starting hyperbola with asymptotes $x=3$ and $y=0$ .

Graph at the selected step

Ideas to try

Which transformation changes the vertical asymptote? (None.)
Which one only changes the horizontal asymptote?

(a)ii

Concavity by calculus

Given instead that $a<-6$ , use a non-graphical method to determine the range of values of $x$ where the graph of $y=\dfrac{2x+a}{3-x}$ is concave upwards.

Hint

Differentiate twice using $y=-2-(6+a)(x-3)^{-1}$ .
Concave upwards means $\frac{d^2y}{dx^2}>0$ .
Use $a<-6$ to decide the sign of $6+a$ .

Answer · (a)ii

x>3

(b)

Reconstructing a graph

The diagrams show the graphs of $y=f(|x|)$ and $y=|f(x)|$ , with asymptotes $x=\pm2$ and $y=2$ marked, and a touching point at $(\pm4,0)$ . The gradient of $y=f(x)$ at the origin is $-\frac34$ . Sketch the graph of $y=f(x)$ , including the points where it crosses the axes and the equations of any asymptotes.

Explore this idea

Hint

For $x\ge0$ , $f(|x|)=f(x)$ , so the right half of the first diagram is the right half of $f$ .
For $x<0$ , compare with $|f(x)|$ . The gradient at the origin is negative, so just to the left $f$ is positive and $|f|=f$ .
Combine the two halves.

Q7 reconstructed f(x)Scroll across to see the full diagram.

Answer · (b)

(0,0)

Explore the idea

Rebuild f, then recreate the two given graphs

Start from the reconstructed f. Applying |x| and then the modulus must give back the two printed diagrams.

Start: y = f(x) (reconstructed)

Through $O$ with slope $-\frac34$ , touching the axis at $(4,0)$ ; asymptotes $x=2$ and $y=2$ .

Marked points at this step, rounded to 6 significant figures.
Point	x	y
O	0	0
(4,0)	4	0

Graph at the selected step

Ideas to try

Compare $f(|x|)$ and $|f(x)|$ on the left. Why are they different there?
Which part of $f$ got flipped by the modulus?

Reconstruct signs from two absolute-value sketches

Solution

Describing transformations

Rewrite the target

Reflect in the $x$ -axis

Stretch parallel to the $y$ -axis by $6+a$

Translate 2 units down

Reflect, stretch, shift

Concavity by calculus

First derivative

Second derivative

Use the sign of $6+a$

Solve the inequality

Reconstructing a graph

Right half from $f(|x|)$

Left half from $|f(x)|$

Check the signs

Rebuild f, then recreate the two given graphs

Rewrite the target

Reflect in the xxx-axis

Stretch parallel to the yyy-axis by 6+a6+a6+a

Translate 2 units down

First derivative

Second derivative

Use the sign of 6+a6+a6+a

Solve the inequality

Right half from f(∣x∣)f(|x|)f(∣x∣)

Left half from ∣f(x)∣|f(x)|∣f(x)∣

Check the signs

Reflect in the $x$ -axis

Stretch parallel to the $y$ -axis by $6+a$

Use the sign of $6+a$

Right half from $f(|x|)$

Left half from $|f(x)|$