Graph 𝑔 cuts the 𝑥-axis at 𝐴(− 3; 0), the 𝑦-axis at 𝐸 and the turning point is at 𝐷. Graph 𝑓 cuts the 𝑦-axis at 𝐶(0; 𝑦) and the asymptote of 𝑓 is 𝑦 =− 3.

Question

Figure 1: Diagram for question 5.
5.1
Show that the equation for 𝑓 is 𝑓(𝑥) = 3. 2𝑥−2 − 3
(3)
5.2
Calculate the 𝑦-coordinate of the point 𝐶.
(1)
5.3
Determine the average gradient between 𝐵 and 𝐶.
(2)
6

5.4
A new function 𝑔 is obtained when 𝑓 is reflected about its asymptote. Determine the equation of 𝑔.
(2)
5.5
Write down the range of 𝑓.
(2)
5.6
Write down the coordinates of 𝐷.
(4)
5.7
Calculate the length of 𝐴𝐹.
(4)
5.8
Determine the value of 𝑘 for which the straight-line graph 𝑦 = 𝑥 + 𝑘 will be a tangent to 𝑔.
(4)

Answer 1

5.1

To find the equation for f, we know that the asymptote is y = -3. This tells us that as x approaches positive or negative infinity, f(x) approaches -3.

Therefore, we can write the equation of f as:

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

5.2

To find the y-coordinate of point C, we substitute x = 0 into the equation for f:

f(0) = 3/(2(0)-2) - 3
f(0) = 3/(-2) - 3
f(0) = -1.5 - 3
f(0) = -4.5

So, the y-coordinate of point C is -4.5.

5.3

To calculate the average gradient between B and C, we can use the formula:

Average gradient = (change in y) / (change in x)

Given that B is (-3, 0) and C is (0, -4.5), the change in y is -4.5 - 0 = -4.5 and the change in x is 0 - (-3) = 3.

Therefore, the average gradient between B and C is:

Average gradient = (-4.5) / 3 = -1.5

5.4

When f is reflected about its asymptote, the equation of g will be the negative reciprocal of f. Therefore, the equation of g will be:

g(x) = -1/(3/(2x-2) - 3)

5.5

The range of f is the set of all possible values that the function can take. Since the asymptote is y = -3, the range of f is all real numbers except -3. So, the range of f is (-∞, -3) U (-3, ∞).

5.6

The turning point D is the point on the graph of f where the gradient is zero. This means that the derivative of f will be zero at D.

To find the coordinates of D, we need to find where the derivative of f equals zero. Differentiating f(x) = 3/(2x-2) - 3 with respect to x, we get:

f'(x) = -6/(2x-2)^2

Setting f'(x) = 0, we get:

-6/(2x-2)^2 = 0
2x-2 = 0
2x = 2
x = 1

Substitute x = 1 back into the original equation for f to find the y-coordinate of D:

f(1) = 3/(2(1)-2) - 3
f(1) = 3/0 - 3 (which is undefined)

Therefore, there is no y-coordinate at the turning point D.

5.7

To calculate the length of AF, we need to find the distance between A (-3, 0) and the x-intercept of f, which occurs when f(x) = 0.

Setting f(x) = 0, we get:

3/(2x-2) - 3 = 0
3/(2x-2) = 3
2x-2 = 1
2x = 3
x = 1.5

Therefore, the x-intercept of f is at (1.5, 0). We can now use the distance formula to find the length of AF:

Length of AF = √((-3 - 1.5)^2 + (0 - 0)^2)
Length of AF = √((-4.5)^2 + 0)
Length of AF = 4.5

5.8

To find the value of k for which the straight-line graph y = x + k will be a tangent to g, we need to find the point of tangency.

Since the gradient of the tangent will be equal to the gradient of g at the point of tangency, we differentiate g (which we found in 5.4) and set it equal to 1 (the gradient of y = x + k).

Differentiating g(x) = -1/(3/(2x-2) - 3), we get:

g'(x) = 3/(2x-2)^2

Setting g'(x) = 1, we get:

3/(2x-2)^2 = 1
2x-2 = √3
2x = √3 + 2
x = (√3 + 2) / 2

Substitute x = (√3 + 2) / 2 back into g to find the y-coordinate of the point of tangency:

g(√3 + 2 / 2) = -1/(3/(2*(√3 + 2) / 2 - 2) - 3)

The value of k is equal to the y-coordinate of the point of tangency.