What is the stationary point and its nature for the function f(x) = x^2-8x+4?

[looi76]

Question:
For the graph x^2-8x+4 :
(i) find the coordinates of the stationary point;
(ii) say, with reasoning, weather this is a maximum or minimum point;
(iii) check your answer by using the method of 'completing the square' to find the vertex;
(iv) state the range of values which the function can take.

Attempt:
(i) f(x) = x^2-8x+4
f'(x) = 2x-8
f'(x) = 0
2x-8 = 0
2x = 8
x = \frac{8}{2} = 4

f(x) = x^2-8x+4
f(x) = 4^2-8\times4+4
f(x) = -12

Stationary Point = (4,-12)

(ii) f'(1) = 2x-8
f'(1) = 2(1)-8
f'(1) = -6

f'(7) = 2x-8
f'(7) = 2(7)-8
f'(7) = 6

x \leq 4
Minimum Point

(iii) Need help...

 

[Kittel Knight]

Here is not the place for homework questions...

(read the first thread)