Unit 7

5 marks · Easy

The temperature, °C, in a greenhouse at time hours is modelled by the cubic function .

(a)

Find the temperature at hours.

[1 mark]

(b)

State the value of (hours) when the graph of cuts the vertical (temperature) axis.

[1 mark]

(c)

The temperature reaches a minimum at some value of . Write down a quadratic equation that must be solved to find the value of where the minimum occurs.

[1 mark]

(d)

Solve the equation using the quadratic formula. Give both values for , correct to 2 decimal places.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) 6 °C

(b) 0 hours

(d) x = 3 + i\sqrt{2},\ x = 3 - i\sqrt{2} \ (complex\ roots) \text{ OR } x = 3.00 \pm 1.41\ (real\ working) · must show Both x values, Correct to 2 decimal places

Common mistakes: Incorrect substitution for T(2); Setting x=0 incorrectly for y-intercept; Incorrect quadratic for minimum; Quadratic formula errors

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Substitute into the cubic and evaluate correctly.

B1 (b) · 1 mark · State or deduce that the temperature axis is cut at .

M1 (c) · 1 mark · Sets the derivative to zero and gives the correct quadratic equation.

M1 (d) · 1 mark · Substitute coefficients into the quadratic formula correctly and attempt to solve.

A1 (d) · 1 mark · Obtain both values for , correct to 2 decimal places. (follow-through)

Worked solution

a) Substitute into :

Correction: 8 - 24 = -16; -16 + 22 = 6; 6 + 2 = 8. So, .

b) The vertical axis (temperature axis) is crossed when .

So, .

c) To find the minimum, set the derivative to zero:

So, quadratic equation: .

d) Solve using the quadratic formula:

So, , (to 2 decimal places).