quadratic function?

Question

the quadratic function f is defined by f(x) = x^2 + 6x + 11

1) express f(x) in the form (x+p)^2 + q, wher p and q are integers.
2)state the value for x for which f(x) is least
3) solve inequality f(x) is greater than 3
4) state the geometrical transformation which maps the curve y = F(x) onto y = f(x+2)

Answer 1

1)
x^2 + 6x + 11
(x+3)(x+3) + 2 (x*x = x^2, 3*x+3*x=6x, 3*3=9, so 2 added to that = 11)
(x+3)^2 + 2

2)
a quadratic function is like
a*x^2 + bx + c, to find the value of x where f(x) is the least you do -b/2a, which is -6/2 = -3.

3)
make a sketch (or use a graphic calculator)
x^2 + 6x + 11 = 3
x^2 + 6x + 8 = 0
(x + 2)*(x + 4) = 0 (as x*x = x^2, 2x+4x=6x, 2*4=8)
this means either x + 2 = 0 or x + 4 = 0 ( in a*b = 0, either a or b has to be zero).
x = -2 or x = -4
Now look at the sketch at -2 and -4, you'll see that any value lower than -4 has an f(x) higher than 3, and everything over -2.

So x < -4 or x > -2

4)
When you take the example of
f(x) = x
f(x) = x + 2
you will see that when you take x=0, the second one will act exactly the same as the first, but it's 'developed' further already. Basically the Y takes a value that would 'normally' occur 2 x's later. So the graph moves 2 to the left, (-2,0).

Answer 2

1) complete the square
x^2 + 6x + 11 = x^2 + 6x + 9 + 2
(x^2 + 6x + 9) + 2
The portion inside the bracket is a square function.

2) differentiate the function.
f'(x) = 2x + 6
A critical point (e.g., max or min) is where f'(x) = 0