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-

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!