1. ## Differentiation Question...

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...

2. 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...
$\displaystyle x^2-8x+4 = (x^2 - 8x + 16) - 16 + 4$ $\displaystyle = (x - 4)^2 - 16 + 4 = (x -$ 4$\displaystyle )^2$ - 12. The quadratic is now in turning point form. By inspection, the turning point is at (4, -12).