f.4 maths (functions&grahs)

10.
(a)
The axis of symmetry : x = -2
Hence, M is the mid-point of OA :
(p + 0)/2 = -2
Hence, p = -4

P(-4, 0) lies on the parabola.
Put x = -4 and y = 0 into the equation y = x² + bx :
0 = (-4)² + b(-4)
4b = 16
b = 4

(b)
Let (-2, a) the coordinates of vertex.

Put x = -2 and y = a into the equation y = x² + 4x :
a = (-2)² + 4(-2)
a = -4

Hence, the coordinates of the vertex = (-2, -4)


6.
(a)
ΔABC is an isosceles Δ with AB = AC
Hence, the height AD bisects BC.
Then, BD = DC = (10/2) cm = 5 cm

Let QD = y cm

∠PQB = ∠ADB = 90° (given)
∠B = ∠B (common ∠)
ΔPQB ~ ΔADB (AAA)
BQ/PQ = BD/AD
(5 - y)/x = 5/20
5 - y = x/4
y = 5 - 0.25x

QR = 2y cm
QR = 2(5 - 0.25x) cm
QR = 10 - 0.5x cm

Area of PQRS
= PQ•QR
= x(10 - 0.5x) cm²
= 10x - 0.5x² cm²

(b)
Area of PQRS (in cm²)
= 10x - 0.5x²
= -0.5[x² - 20x]
= -0.5[x² - 20x + (20/2)²] + 0.5(20/2)²
= -0.5(x - 10)² + 50

Since -0.5(x - 10)² ≤ 0
Then, area of PQRS ≤ 50 cm²

When x = 10, the max. area of PQRS = 50 cm²

When x = 10 : QD = 10 - 0.5(10) cm = 5 cm
Dimension of PQRS = 5 cm x 10 cm