the-curve-y-f-x-when-f-x-is-a-quadratic-expression-has-a-maximum-value-point-at-1-4-The-curve-touches-the-line-6x-y-13-Find-the-value-of-x-for-which-y-8-

Question Number 71196 by Rio Michael last updated on 12/Oct/19

t h e c u r v e y = f (x), w h e n f (x) i s a q u a d r a t i c e x p r e s s i o n h a s

a m a x i m u m v a l u e p o i n t a t (1, 4) . T h e c u r v e t o u c h e s t h e l i n e

6 x + y = 13 . F i n d t h e v a l u e o f x f o r w h i c h y = 8

Answered by MJS last updated on 12/Oct/19

f (x) is quadratic \Leftrightarrow y = {a x}^{2} + b x + c

(\begin{matrix} 1 \\ 4 \end{matrix}) \in f (x) \Leftrightarrow 4 = a + b + c \Rightarrow c = 4 - a - b

\Rightarrow y = {a x}^{2} + b x + 4 - a - b

maximum at (\begin{matrix} 1 \\ 4 \end{matrix}) \Leftrightarrow f^{'} (1) = 0 \land f^{″} (1) < 0

y^{'} = 2 a x + b \to 2 a + b = 0 \Rightarrow b = - 2 a

y^{″} = 2 a \to a < 0

\Rightarrow y = {a x}^{2} - 2 a x + a + 4

y = 13 - 6 x is tangent

tangent in (\begin{matrix} p \\ f (p) \end{matrix}) \in f (x) : y = 2 a (p - 1) x + a (1 - p^{2}) + 4

\Rightarrow 2 a (p - 1) = - 6 \land a (1 - p^{2}) + 4 = 13

\Rightarrow a = - 3 \land p = 2

\Rightarrow y = - 3 x^{2} + 6 x + 1

y = 8 \Rightarrow 8 = - 3 x^{2} + 6 x + 1 \Rightarrow x = 1 \pm \frac{2 \sqrt{3}}{3} i

Commented by Rio Michael last updated on 13/Oct/19

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