Find the maximum or minimum value of each quadratic function:?

Question

1. f(x)=x^2-5x+6
2. f(x)=x^2-10x+21
3. f(x)=x^2-x-6
4. f(x)=x^2-2x+4
5. f(x)=2x^2-x+6
6. f(x)=6x^2-x+2
7. f(x)=x^2+4x-5
8. f(x)=2x^2+2x-4
9. f(x)= -x^2+4x-3
10. f(x)= -x^2-5x-3
pls answer with complete solution...

Answer 1

I think you should do your own homework, so I'm going to do 1, and show you how.

To find the minimum or maximum, differentiate the function and solve for slope =0.

f(x) = x² -5x +6
f'(x) = 2x -5

so if 2x -5 = 0, then
x =2.5

f(2.5) = (2.5)² - 5(2.5) + 6
f (2.5) = 6.25 - 12.5 + 6
f(2.5) = -0.25 is the minimum.

Answer 2

The first derivative test will tell you the maximum and minimum points. The second derivative test will tell you whether the graph is concave up or down at the critical points. Concave up means you have a minimum, concave down means you have a maximum.

For example,

1.)
f = x^2 - 5x + 6
f' = 2x - 5
f'' = 2

when f' = 0, you have found either a maximum or minimum value.

since f'' = 2, the graph is concave up.

For this problem, when x = 2.5, you are at the minimum point.

f(2.5) = 24.75

so the minimum point is (2.5, 24.75)

Answer 3

differentiate and set the differential to 0, then solve for x.

eg

x^2-5x + 6

dy/dx = 2x - 5

x = -5/2

then set x to a value a bit each side of your solution, work out the gredients each side and you will see if it is a max or a min. believe me you are better off with help than with a complete solution though you may not think it :)

Answer 4

Not going to do all of these. Take the first derivative, set it equal to zero and solve for x.
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