given that f(x)=x^2-4x-5, state the coordinates of the turning point?

Answer 1

Consider this:

let y = x^2 - 4x - 5, so

dy/dx = 2x - 4.

At any turning point, dy/dx = 0, so

2x - 4 = 0, so

x = 2.

When x = 2,

y = 2^2 -8 -5,

y = 4 - 8 - 5

y = -9.

d^2y/dx^2 = 2, so the turning point is a minimum.

Hence the turning point of y = x^2-4x-5 is (2, -9)

Hope this helps, Twiggy.

Answer 2

F=(x+1)(x-5)

Answer 3

f(x)=x^2-4x-5
x- c00rdinate of vertex is -b/2a = 4/2=2
y-coordinate is 2^2-4*2-5 = -9
So vertex is (2,-9) which is the turning point