WMA11 June 2022

• WMA11/01​​ Edexcel​​ IAL P1 June 2022, Q5 (Quadratics &​​ Straight Line Graphs)

The curve C has equation y = f(x)​​ 

Given that​​ 

• f(x) is a quadratic expression​​ 

• the maximum turning point on C has coordinates (−2, 12)​​ 

• C cuts the negative x-axis at −5​​ 

• find f(x)​​ 

(4)

The line​​ l1​​ has equation​​ y =45x.

Given that the line​​ l2​​ is perpendicular to​​ l1,​​ and passes through (−5, 0)​​ 

(b) find an equation for​​ l2​​ , writing your answer in the form y = mx + c where m and c are constants to be found.​​ 

(3)

Figure 2 shows a sketch of the curve C and the lines​​ l1​​ and​​ l2​​ 

(c) Define the region​​ R, shown shaded in Figure 2, using inequalities.​​ 

(2)

SOLUTION​​ 

a-​​ 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.

Therefore, using vertex​​ -2, 12 ​​ and the x-intercept (-5, 0) to find the function expression.​​ 

y=a x-p2+q

Where​​ (-p, q) is the coordinate of vertex or turmning point.​​ 

y=ax+22+12

Sustituting the x-intercept point​​ -5, 0.

0=a-5+22+12

-12=a-32

-12=9a 

a=-129 

a=-43

Hence, on substituting the value of a in the equation.

fx=-43 x+22+12

Fx=-43 x2+4x+4+12

fx=-43x2-163x-163+12

fx=-43x2-163x-163+363

fx= -43x2-163x+203

b-​​ Since​​ l1​​ and​​ l2​​ are perpendicluar and the gradient of​​ l1​​ is​​ 45, the gradient of​​ l2​​ will be​​ 

m1×m2=-1

45×m2=-1

m2= -54

Now, using point slope formula​​ where slope of​​ l2​​ is -5/4 and point is​​ (−5, 0).

y-y1=mx-x1

y-0=-54 x--5 

y= -54 x+3

y= -54x-254

c-​​ 

Since the region is enclosed between the solid lines, the inequalities must be either​​ ≤​​ or​​ ≥.​​ 

Keeping in view the above flash card.​​ 

Region is​​ 

• below/inside the quadratic curve; therefore,

y≤43x2-163x+203

• above the straight line,​​ l1; therefore,​​ 

y≥45 x 

• above line​​ l2, so​​ 

y≥ -54x -254