Expert: Scott A Wilson - 3/20/2010

Question
These challange me for some reason :(
Please help!

1. Find the min/max and vertex of the function
  g(x) = 2x^2 + 8x


2. Find the min/max and vertex of the function:
   h(x) = (2x - 5)(5 + x)


3. Find the min/max and vertex of the function:
   g(x) = 2x^2 - 6x^2 + 2


4. Find a quadratic equation with integral coefficients having  the given roots:
4 + 2i, 4 - 2i


5. Find a quadratic equation with integral coefficients having the given roots:
i√ 5 , -i√ 5

Answer
1. Find the min/max and vertex of the function g(x) = 2x² + 8x.
This function has g'(x) = 2x + 8, and g"(x) = 2.
It is known if the 2nd derivative is +, the function is concave up.
This means there is a minimum where the vertex is at, and that is where g'(x) = 0.
There is no maximum.

2. Find the min/max and vertex of the function h(x) = (2x - 5)(5 + x).
When multiplied out the equation is h(x) = 2x² + 5x - 25.
It can be seen that h'(x) = 4x + 5 and h"(x) = 4.
f">0 => concave up; h'=0 at x = -5/4; h(-5/4) is a minimum; there is no max.


3. Find the min/max and vertex of the function g(x) = 2x² - 6x^2 + 2;
is that right, -6x has a ² after the x?

If that is true, note g(x) = -4x² + 2, so g'(x) = -8x and g"(x) = -8.
Since g"(x) < 0, the critical point is a max.
Since -8x = 0 at x = 0, that is where the critical point is at.

4. Find a quadratic equation with integral coefficients having the given roots 4 + 2i, 4 - 2i.
To do this, take (x - 4 - 2i)(x - 4 + 2i).

Note that this is of the form (a-b)(a+b) where a is x-4 and b is 2i.
The answer to this is a² - b² = (x-4)² - (2i)².
It is known that (x-4)² = x² - 8x + 16 and (2i)² = -4, so (x-4)² - (2i)² = x² - 8x + 12.

5. Find a quadratic equation with integral coefficients having the given roots: i√5 , -i√5.
That would be (x-i√5)(x+i√5) = x² - (i√5)² = x² + 25.