Question

symmetry of F(x)= x^2-4/x^2-2x

Answer 1

f(x)=x²+2x-8 is in the form ax² + bx + x

To find the x intercepts, factorise the equation and let y = 0:
x² + 2x - 8
= (x + 4)(x - 2) = 0
x intercepts = -4 and 2

To find y intercepts let x = 0.
x² + 2x - 8
0² + 2(0) - 8
= -8
y intercept = -8

The formula for axis of symmetry is -b/2a.
b = 2, a = 1
-2/2 = -1
axis of symmetry = -1

Vertex occurs at the axis of symmetry. The axis of symmetry is the x coordinate of the vertex. To find the y coordinate, sub the value of the axis of symmetry into the equation.

axis of symmetry = -1
y = x² + 2x - 8
y = -1² + 2(-1) - 8
y = 1 - 2 - 8
y = -9

turning point = (-1, -9)

The domain is the values of x you can put into the equation.
For this, every real number can be put in the equation, so the domain is R (all real numbers), expressed as:
Domain = R

The range are the y values that come out from putting any x values into the equation. The vertex is the lowest y value produced.
The graph goes from -9 (the y coordinate of the turning point) upto infinity.
The range is expressed as:
Range = [-9, ∞)

Answer 2

hope this helps ;o

Answer 3

you didnt use my function tho...

Answer 4

this was suppose to give you an example .. but hold on ill answer it for you (:

Answer 5

ok thanks