AS .. MATHS Unit Pure Core1

1i)
Gradient of AB = (-1-5) / (2-6) = (-6)/(-4) = 3/2

1ii)
Since the gradient of AB is 3/2, the equation of AB is:
(y-5) / (x-6) = 3/2
2(y-5) = 3(x-6)
2y-10 = 3x-18
3x-2y = 8


2a)
y = x² - 6x + 16
y = (x² - 6x + 9) - 9 + 16 【Method of completing the square】
y = (x - 3)² + 7
So, the eqaution of the line of symmetry of the curve is x = 3.

2b)
The transformation that maps y = x² to y = x² - 6x + 16 = (x - 3)² + 7 is a transformation that moves the graph of y = x² to the right by 3 units, then move it up by 7 units.


3.
(x-2)(x²+x+3)=0
x - 2 = 0 or x²+x+3 = 0
Consider the discriminant of x²+x+3.
Δ = 1² - 4(1)(3) = -11＜0
So the equation x²+x+3 = 0 has no real roots.
Therefore, x²+x+3 = 0 is rejected and so the equation has only 1 real root.

The value of the real root is:
x - 2 = 0
x = 2


Hope it helps! ^^

1)
Since, y = m x + c
5 = m 6 + c .... (1)
-1 = m 2 + c ....(2)
(1) - (2) 6 = 4 m
m is the gradient or the slope
m = 3 / 2 or 1.5
5 = 6 * 3 / 2 + c
c = - 4
y = 3 / 2 x - 4
3x - 2y = 8

2)
y = x^2 - 6x + 16
y - 7 = x^2 - 6x + 9
(y - 7) = (x - 3)^2
When compare with the curve y' = x' ^2
y' = y-7 & x' = x-3, so the curve is transform upward and right from (0,0) to (3,7)
The equation of the line of symmetry of the cure is "x = 3"