2014 PAPER 1 – QUESTION 5.

The function f is defined as

⟼ ∈ f:x x 3+3 x 2–9 x + 5, where x R. (a)(i) Find the co-ordinates of the point where the graph of f cuts the y-axis.

All functions cross the y-axis when x =0. Substitute this into the function above. f(x )= x 3+3 x 2–9 x +5 f(0) = (0)3 + 3(0)2 – 9(0) + 5 = 5 Thus, the y-intercept is (0, 5).

(ii) Verify that the graph of f cuts the x-axis at x =–5.

A graph cuts the x-axis when y = 0. In this question, substitute x = – 5 into the function. This will result in the answer being zero.

3 2

f(–5) = (–5) + 3(–5) – 9 (–5) +5

=–125+75+45+5=0

⇒

f(–5) = 0 (–5,0) is the x-intercept.

(b) Find the co-ordinates of the local maximum turning point and the local minimum turning point of f.

The formula to use here is: f’(x) = 0 (this is the same as = 0). This formula is not in the Maths tables, this making it a formula that students must learn. f’( x) =0 f’( x) = 3x 2+6 x –9 f’( x) = 3x 2+6 x –9=0

Factorising the above quadratic equation yields:

⇒(3x

+ 9)(x –1)=0

⇒

3x +9=0 x –1=0

⇒

3x =–9 x =1

⇒

x = –3 and x =1

Substitute these values of x into f(x) in order to get the y-value of the co-ordinates. x =1 : f(1)= (1)3 + 3(1)2 – 9(1) + 5

=1+3–9+5

∴

=0 (1,0) x =–3:

f(–3) = (–3)3 + 3(–3)2 – 9(–3) + 5

=–27+27+27+5

∴

=32 (-3,32)

The turning point with the larger y-value is the maximum turning point. Since 32 > 0, the point (-3,32) is the maximum turning point. Subsequently, (1,0) is the minimum turning point.

(c) Hence, sketch the graph of the function f on the axes below.

The word ‘hence’ indicates that you must use parts (a) and (b) to answer part (c).