Question

Maths functions question

Answer 1

a) OA = 1 unit

b) BC = 3 units

c) OD = 2 units

d) AC = 3√2 units

Explanation:

Given function:

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

Part (a)

Point A is the x-intercept of the curve.

To find the x-intercept of the curve (when y = 0), set the function to zero and solve for x:

`\begin{aligned}f(x) & = 0\\\implies (2)/(x)-2 & = 0\\(2)/(x) & = 2\\2 & = 2x\\\implies x & = 1\end{aligned}`

Therefore, A (1, 0) and so OA = 1 unit.

Part (b)

If OB = 2 units then B (-2, 0). Therefore, the x-value of Point C is x = -2.

To find the y-value of Point C, substitute x = -2 into the function:

`\implies f(-2)=(2)/(-2)-2=-3`

Therefore, C (-2, -3) and so BC = 3 units.

Part (c)

Asymptote: a line that the curve gets infinitely close to, but never touches.

The y-value of Point D is the horizontal asymptote of the function.

The function is undefined when x = 0 and therefore when y = -2.

Therefore, D (0, -2) and so OD = 2 units.

Part (d)

From parts (a) and (c):

• A = (1, 0)
• C = (-2, -3)

To find the length of AC, use the distance between two points formula:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

`\textsf{where }(x_1,y_1) \textsf{ and }(x_2,y_2)\:\textsf{are the two points.}`

Therefore:

`\sf \implies AC=√((x_C-x_A)^2+(y_C-y_A)^2)`

`\sf \implies AC=√((-2-1)^2+(-3-0)^2)`

`\sf \implies AC=√((-3)^2+(-3)^2)`

`\sf \implies AC=√(9+9)`

`\sf \implies AC=√(18)`

`\sf \implies AC=√(9 \cdot 2)`

`\sf \implies AC=√(9)√(2)`

`\sf \implies AC=3√(2)\:\:units`