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)