Question

by expressing y-4x-x^2+8 in form y=a(x-p)^2+q, where p and q are constant, state
a) the equation of the axis of symmetry
b) the coordinates of the turning point on the curve

Answer 1

can u express y=4x-x^2+8 in the form y=a(x-p)^2+q,

Answer 2

@sauravshakya nope

Answer 3

y=4x-x^2+8
y=-(x^2-4x)+8
y=-(x^2-2*x*2+2^2-2^2)+8
y=-(x-2)^2+8+8
y=-(x-2)^2+16

Answer 4

got it?

Answer 5

hint: for y=(x-h)^2+k
Co-ordinates of turning point===> (h,k)
ANd axis of symmetry is x=h