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


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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

	$t h a n k s s i r$
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