Let function f(x) be defined as f(x)= x^2 +bx+c , where b,c∈R . And f(1) − 2f(5) +f(9) =32. Find no. of ordered pairs (b,c) such that ∣f(x)∣≤8 ∀ x∈ [1,9] ?


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Question Number 33651 by rahul 19 last updated on 21/Apr/18

$L e t f u n c t i o n f (x) b e d e f i n e d a s$ $f (x) = x^{2} + b x + c, w h e r e b, c \in R .$ $A n d f (1) - 2 f (5) + f (9) = 32 .$ $F i n d n o . o f o r d e r e d p a i r s (b, c)$ $s u c h t h a t ∣ f (x) ∣⩽ 8 \forall x \in [1, 9] ?$
	Answered by MJS last updated on 21/Apr/18

	$something about polynomial functions of$ $2^{nd} degree$ $there are 2 different points of view$ $usually we are focused on zeros$ $x^{2} + p x + q = 0 \Leftrightarrow$ $\Leftrightarrow (x - x_{1}) (x - x_{2}) \Leftrightarrow$ $\Leftrightarrow p = - (x_{1} + x_{2}) \land q = x_{1} x_{2}$ $but we should think about steepness, orientation$ $and shifts in both x - and y - directions$ $y = {a x}^{2} + b x + c$ $a > 0 \Leftrightarrow ‘ ‘ {hanging}^{″} parabola$ $a < 0 \Leftrightarrow ‘ ‘ standing {parabola}^{″}$ $∣ a ∣ determines steepness$ $y^{'} = 2 a x + b$ $∣ 2 a ∣< 1 \Leftrightarrow ‘ ‘ {flat}^{″} or ‘ ‘ {wide}^{″} parabola$ $∣ 2 a ∣> 1 \Leftrightarrow ‘ ‘ {steep}^{″} or ‘ ‘ {narrow}^{″} parabola$ $b and c have got something to do with the$ $x - and y - shifts$ ${let}^{'} s look at the basic function$ $y = x^{2}$ $shift up / down$ $y - v = x^{2} \Leftrightarrow y = x^{2} + v$ $shift left / right$ $y = {(x + u)}^{2} + v \Leftrightarrow y = x^{2} + 2 u x + u^{2} + v$ $b = 2 u \land c = u^{2} + v \Leftrightarrow u = \frac{b}{2} \land v = c - \frac{b^{2}}{4}$ $so y = x^{2} + b x + c is a shifted basic parabola .$ $we are looking for a part of this function$ $which fits into a window of width 8 and$ $height 16 (because the width of the given$ $interval [1; 9] is 8 and abs (f (x)) ⩽ 8 \Leftrightarrow$ $\Leftrightarrow - 8 ⩽ f (x) ⩽ 8 \Leftrightarrow 0 ⩽ f (x) + 8 ⩽ 16$ $we can look at y = x^{2} and afterwards do the$ $shifting$ $1 . try on left arm (symm . to right arm)$ $f (x) : y = x^{2}$ $x ⩽ 0$ $f (x) \pm 16 = f (x - 8)$ $x^{2} \pm 16 = x^{2} - 16 x + 64$ $16 x = 64 \pm 16$ $[x = 3 \lor x = 5 \Rightarrow no solution on left arm$ $f (\pm 3) = 9; f (\pm 5) = 25 but f (0) = 0 \Rightarrow our$ $window would be 8 \times 25 instead of 8 \times 16]$ $2 . try center$ $f (- 4) = 16$ $f (0) = 0$ $f (4) = 16$ $is the only solution$ $but we need x \in [1; 9] instead of x \in [- 4; 4] \Rightarrow$ $\Rightarrow shift 5 to the right$ $y = {(x - 5)}^{2}$ $and we need y \in [- 8; 8] instead of y \in [0; 16] \Rightarrow$ $\Rightarrow shift 8 down$ $y = {(x - 5)}^{2} - 8$ $y = x^{2} - 10 x + 17$ ${there}^{'} s only one pair (b; c) = (- 10; 17)$ $[f (1) - 2 f (5) + f (9) = 32 is true for any pair$ $(b; c), so we {didn}^{'} t need it :$ $f (1) = 1 + b + c$ $f (5) = 25 + 5 b + c$ $f (9) = 81 + 9 b + c$ $(1 + b + c) - 2 (25 + 5 b + c) + (81 + 9 b + c) = 32]$
	Commented by rahul 19 last updated on 22/Apr/18

	$t h a n k u s o m u c h s i r!$
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