1. ## Differentiation Question...

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...

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