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

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

$${the}\:{curve}\:{y}\:=\:{f}\left({x}\right),\:{when}\:{f}\left({x}\right)\:{is}\:{a}\:{quadratic}\:{expression}\:{has}\: \\ $$$${a}\:{maximum}\:{value}\:{point}\:{at}\:\left(\mathrm{1},\mathrm{4}\right).\:{The}\:{curve}\:{touches}\:{the}\:{line} \\ $$$$\mathrm{6}{x}\:+\:{y}\:=\:\mathrm{13}.\:{Find}\:{the}\:{value}\:{of}\:{x}\:{for}\:{which}\:{y}\:=\:\mathrm{8} \\ $$

Answered by MJS last updated on 12/Oct/19

f(x) is quadratic ⇔ y=ax^2 +bx+c ((1),(4) )∈f(x) ⇔ 4=a+b+c ⇒ c=4−a−b ⇒ y=ax^2 +bx+4−a−b maximum at ((1),(4) ) ⇔ f′(1)=0 ∧ f′′(1)<0 y′=2ax+b → 2a+b=0 ⇒ b=−2a y′′=2a → a<0 ⇒ y=ax^2 −2ax+a+4 y=13−6x is tangent tangent in ((p),((f(p))) )∈f(x): y=2a(p−1)x+a(1−p^2 )+4 ⇒ 2a(p−1)=−6∧a(1−p^2 )+4=13 ⇒ a=−3∧p=2 ⇒ y=−3x^2 +6x+1 y=8 ⇒ 8=−3x^2 +6x+1 ⇒ x=1±((2(√3))/3)i

$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{quadratic}\:\Leftrightarrow\:{y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\begin{pmatrix}{\mathrm{1}}\\{\mathrm{4}}\end{pmatrix}\in{f}\left({x}\right)\:\Leftrightarrow\:\mathrm{4}={a}+{b}+{c}\:\Rightarrow\:{c}=\mathrm{4}−{a}−{b} \\ $$$$\Rightarrow\:{y}={ax}^{\mathrm{2}} +{bx}+\mathrm{4}−{a}−{b} \\ $$$$\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{4}}\end{pmatrix}\:\Leftrightarrow\:{f}'\left(\mathrm{1}\right)=\mathrm{0}\:\wedge\:{f}''\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${y}'=\mathrm{2}{ax}+{b}\:\rightarrow\:\mathrm{2}{a}+{b}=\mathrm{0}\:\Rightarrow\:{b}=−\mathrm{2}{a} \\ $$$${y}''=\mathrm{2}{a}\:\rightarrow\:{a}<\mathrm{0} \\ $$$$\Rightarrow\:{y}={ax}^{\mathrm{2}} −\mathrm{2}{ax}+{a}+\mathrm{4} \\ $$$${y}=\mathrm{13}−\mathrm{6}{x}\:\mathrm{is}\:\mathrm{tangent} \\ $$$$\mathrm{tangent}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix}\in{f}\left({x}\right):\:{y}=\mathrm{2}{a}\left({p}−\mathrm{1}\right){x}+{a}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)+\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2}{a}\left({p}−\mathrm{1}\right)=−\mathrm{6}\wedge{a}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)+\mathrm{4}=\mathrm{13} \\ $$$$\Rightarrow\:{a}=−\mathrm{3}\wedge{p}=\mathrm{2} \\ $$$$\Rightarrow\:{y}=−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1} \\ $$$${y}=\mathrm{8}\:\Rightarrow\:\mathrm{8}=−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\pm\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{i} \\ $$

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

thanks sir

$${thanks}\:{sir} \\ $$

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