WMA11 October 2021

8.​​ WMA11/01 Edexcel​​ IAL P1 Oct 2021 Q8​​ (Quadratics)

A graph of a function

Description automatically generated

Figure 4 shows a sketch of the curve C with equation​​ 

y = 4 + 12x  3x2 

The point M is the maximum turning point on C.​​ 

(a)

(i) Write​​ 4 + 12x  3x2​​ in the form​​ 

a + bx + c2 

where a, b and c are constants to be found.​​ 

(ii) Hence, or otherwise, state the coordinates of M.​​ 

(5)

The line l1​​ passes through O and M, as shown in Figure 4.​​ 

A line l2​​ touches C and is parallel to l1​​ 

(b) Find an equation for l2​​ 

(5)

SOLUTION

a-​​ 

i-​​ Using completing square method to express the quadratic equation in the given form.

 4+12x-3x2=-3x2+12x+4

=-3x2-4x+4

=-3x-22-4+4 

=-3x-22+12+4

=-3 x-22+16

=16-3x-22

ii-​​ 

The quadratic equation can be expressed in the form of completing square form as​​ y=ax-p2+q, where (-p, q) is the vertex of the quardratic function.

 

In this question, the vertex of a quadratic equation is given as (2, 16).

M2, 16 

   

b-​​ 

Since l1​​ and l2​​ are parallel, there gradients are equal to each other. So, first we will find the gradient of line l1.​​ 

The gradient of l1​​ as it passes through O (0,0) & M (2,16).

m1=y2-y1x2-x1

m1=16-02-0 

m1=8

So, the gradient of line l2​​ is​​ 

m1=m2=8

Now, the equation of line l2​​ is​​ 

y=8x+c

As it is given that line​​ l2​​ touches C, it means the line l2​​ cuts curve at a single point, or in ither words, line l2​​ is tangential to the curve. Hence, solving both equations simultaneously.

8x+c=4+12x-3x2

3x2-4x+c-4= 0

Since the line touches curve or is tangential, so the above simultaneously solved equation must have one solution, which means the discriminant is equal to 0.​​ 

b2-4ac=0

-42-43c-4=0

16-12c+48=0

64=12c

64=12c

c=162

Putting this value of c in the equation we made earlier of line l2.

y=8x+163

3y=24x +16

24x-3y+16=0