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Question Number 83964 by Rio Michael last updated on 08/Mar/20

The graph of y = ((a + bx)/((x−1)(x−4))) has a turning point at P(2,−1). Find the value of a and b and hence,sketch the curve y = f(x) showing clearly the turning points, asympototes and intercept(s) with the axes.

$$\mathrm{The}\:\mathrm{graph}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\:=\:\frac{{a}\:+\:{bx}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{4}\right)} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{turning}\:\mathrm{point}\:\mathrm{at}\:{P}\left(\mathrm{2},−\mathrm{1}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b}\: \\ $$$$\mathrm{and}\:\mathrm{hence},\mathrm{sketch}\:\mathrm{the}\:\mathrm{curve}\:{y}\:=\:{f}\left({x}\right)\:\mathrm{showing}\:\mathrm{clearly}\:\mathrm{the} \\ $$$$\mathrm{turning}\:\mathrm{points},\:\mathrm{asympototes}\:\mathrm{and}\:\mathrm{intercept}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{axes}. \\ $$

Answered by john santu last updated on 08/Mar/20

(1) −1 = ((a+2b)/(1.(−2))) ⇒ a+2b = 2 (2)y = ((2−2b+bx)/(x^2 −5x+4)) y ′ = ((b(x^2 −5x+4)−(2x−5)(2−2b+bx))/((x−1)^2 (x−4)^2 )) y ′ = 0 for x = 2 −2b −(−1)(2) = 0 2b = 2 ⇒ b = 1 ∧ a = 0 ∴ y = (x/(x^2 −5x+4)) . vertical asymtote x = 1 ∧ x = 4 horizontal asymtote y = lim_(x→∞) (x/(x^2 −5x+4)) = 0

$$\left(\mathrm{1}\right)\:−\mathrm{1}\:=\:\frac{\mathrm{a}+\mathrm{2b}}{\mathrm{1}.\left(−\mathrm{2}\right)}\:\Rightarrow\:\mathrm{a}+\mathrm{2b}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right){y}\:=\:\frac{\mathrm{2}−\mathrm{2b}+\mathrm{b}{x}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}} \\ $$$${y}\:'\:=\:\frac{\mathrm{b}\left({x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}\right)−\left(\mathrm{2}{x}−\mathrm{5}\right)\left(\mathrm{2}−\mathrm{2}{b}+{bx}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${y}\:'\:=\:\mathrm{0}\:\mathrm{for}\:{x}\:=\:\mathrm{2} \\ $$$$−\mathrm{2}{b}\:−\left(−\mathrm{1}\right)\left(\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{2}{b}\:=\:\mathrm{2}\:\Rightarrow\:\mathrm{b}\:=\:\mathrm{1}\:\wedge\:\mathrm{a}\:=\:\mathrm{0}\: \\ $$$$\therefore\:\mathrm{y}\:=\:\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:. \\ $$$${vertical}\:{asymtote}\:{x}\:=\:\mathrm{1}\:\wedge\:{x}\:=\:\mathrm{4} \\ $$$${horizontal}\:{asymtote}\: \\ $$$${y}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:=\:\mathrm{0} \\ $$$$ \\ $$

Commented by jagoll last updated on 08/Mar/20

Commented by jagoll last updated on 08/Mar/20

this is the graph y = (x/(x^2 −5x+4))

$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{y}\:=\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{4}} \\ $$

Commented by Rio Michael last updated on 08/Mar/20

thanks sirs great work i appreciate

$${thanks}\:{sirs}\:{great}\:{work} \\ $$$${i}\:{appreciate} \\ $$

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