Originally Posted by

**looi76** **Question:**

For the graph $\displaystyle 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) $\displaystyle f(x) = x^2-8x+4$

$\displaystyle f'(x) = 2x-8$

$\displaystyle f'(x) = 0$

$\displaystyle 2x-8 = 0$

$\displaystyle 2x = 8$

$\displaystyle x = \frac{8}{2} = 4$

$\displaystyle f(x) = x^2-8x+4$

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

$\displaystyle f(x) = -12$

Stationary Point = $\displaystyle (4,-12)$

(ii) $\displaystyle f'(1) = 2x-8$

$\displaystyle f'(1) = 2(1)-8$

$\displaystyle f'(1) = -6$

$\displaystyle f'(7) = 2x-8$

$\displaystyle f'(7) = 2(7)-8$

$\displaystyle f'(7) = 6$

$\displaystyle x \leq 4$

Minimum Point

(iii) Need help...