Help!! Function: f(x)= (x^2)-2x-7........?

Question

Function: f(x)= (x^2)-2x-7

1) Find all intercepts of the graph of f (Please include all steps especially when finding x)
2) Express the function f in standard form
3)Find the vertex and axis of symmetry

Answer 1

f(x)= x^2-2x-7 [parentheses unecessary]
Use quadratic formula:
x = [2 +/- sqrt(4+28)]/2
x = 1 +/- 2sqrt(2)
So x intercepts are:
(1+2sqrt(2), 0) and (1-2sqrt(2),0)
The y-intercept is when f(x) = 0 so is (0,-7).

Different people have different ideas as to what the standard for for a parabola is. Some would say it is just x^2-2x-7. Others would say it is (x-1)^2 = y+8 (which is the vertex form). Ohers would give some other form. But in the end,they are all the same.

The vertex is at (1,-8). See vertex form above
The axis of symmetry is x=-b/2a = -(-2)/2 = 1
When f(x)=1 y = y-coordinate of vertex = -8

Answer 2

1) Quadratic formula
x=[2±√([-2]^2-4*1*-7)]/(2*1)
x=[2±√(4-[-28])]/2
x=[2±√(4+28)]/2
x=[2±√(32)]/2
x=[2±4√(2)]/2
x=1±2√(2)
x=3.83 or -1.83 (rounded)

2) y=(x-1)^2-9
3) vertex is (1,-9); symmetric with respet to the y-axis

Answer 3

x - int = using quadratic formula...

(5.657, 0), ( -5.657)

standard form - (x-1)^2 - 8
vertex = (1, - 8)
axis of symmetry = x = 1