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Question Number 194809 by dimentri last updated on 16/Jul/23

If f(x)=ax^2 −5x+3 and g(x)=3x−3 intersection at points (1,h) and (3,t). Find

$$\:\:\:\:{If}\:{f}\left({x}\right)={ax}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\:{and}\: \\ $$$$\:\:\:{g}\left({x}\right)=\mathrm{3}{x}−\mathrm{3}\:{intersection}\:{at} \\ $$$$\:{points}\:\left(\mathrm{1},{h}\right)\:{and}\:\left(\mathrm{3},{t}\right). \\ $$$$\:\:{Find}\:\:\underline{ } \\ $$

Answered by horsebrand11 last updated on 16/Jul/23

f(x)=a(x−1)(x−3)+g(x) ax^2 −5x+3 = ax^2 −4ax+3a+3x−3 ax^2 −5x+3=ax^2 −(4a−3)x+3(a−1) { ((5=4a−3)),((1=a−1)) :}⇒ determinant (((a=2))) ∴ f(x)=2x^2 −5x+3 f(2)−f(−2)=1−21= determinant (((−20)))

$$\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{ax}^{\mathrm{2}} −\mathrm{5x}+\mathrm{3}\:=\:\mathrm{ax}^{\mathrm{2}} −\mathrm{4ax}+\mathrm{3a}+\mathrm{3x}−\mathrm{3} \\ $$$$\:\:\mathrm{ax}^{\mathrm{2}} −\mathrm{5x}+\mathrm{3}=\mathrm{ax}^{\mathrm{2}} −\left(\mathrm{4a}−\mathrm{3}\right)\mathrm{x}+\mathrm{3}\left(\mathrm{a}−\mathrm{1}\right) \\ $$$$\:\:\begin{cases}{\mathrm{5}=\mathrm{4a}−\mathrm{3}}\\{\mathrm{1}=\mathrm{a}−\mathrm{1}}\end{cases}\Rightarrow\begin{array}{|c|}{\mathrm{a}=\mathrm{2}}\\\hline\end{array} \\ $$$$\:\:\therefore\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{3} \\ $$$$\:\:\mathrm{f}\left(\mathrm{2}\right)−\mathrm{f}\left(−\mathrm{2}\right)=\mathrm{1}−\mathrm{21}=\begin{array}{|c|}{−\mathrm{20}}\\\hline\end{array} \\ $$

Commented by dimentri last updated on 16/Jul/23

x

$$\:\:\:\:\:\:\underline{\underbrace{\boldsymbol{{x}}}} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Jul/23

f(1)=g(1)=h a(1)^2 −5(1)+3=3(1)−3=h a−2=0 a=2 f(x)=2x^2 −5x+3 f(2)−f(−2) ={2(2)^2 −5(2)+3}−{2(−2)^2 −5(−2)+3} =1−21=−20 •“(3,t) is intersection point”is unnecessary.One common point is sufficient here.

$${f}\left(\mathrm{1}\right)={g}\left(\mathrm{1}\right)={h} \\ $$$${a}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{1}\right)+\mathrm{3}=\mathrm{3}\left(\mathrm{1}\right)−\mathrm{3}={h} \\ $$$${a}−\mathrm{2}=\mathrm{0} \\ $$$${a}=\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(−\mathrm{2}\right) \\ $$$$\:\:\:\:=\left\{\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{2}\right)+\mathrm{3}\right\}−\left\{\mathrm{2}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{5}\left(−\mathrm{2}\right)+\mathrm{3}\right\} \\ $$$$\:\:\:=\mathrm{1}−\mathrm{21}=−\mathrm{20} \\ $$$$\bullet``\left(\mathrm{3},{t}\right)\:{is}\:{intersection}\:{point}''{is} \\ $$$${unnecessary}.{One}\:{common}\:{point} \\ $$$${is}\:{sufficient}\:{here}. \\ $$

Commented by dimentri last updated on 16/Jul/23

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