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


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

$I f f (x) = {a x}^{2} - 5 x + 3 a n d$ $g (x) = 3 x - 3 i n t e r s e c t i o n a t$ $p o i n t s (1, h) a n d (3, t) .$ $F i n d \underset{―}{}$
	Answered by horsebrand11 last updated on 16/Jul/23
	$f(x)=a(x−1)(x−3)+g(x) ax^2 −5x+3 = ax^2 −4ax+3a+3x−3 ax^2 −5x+3=ax^2 −(4a−3)x+3(a−1) { ((5=4a−3)),((1=a−1)) :}⇒ determinant (((a=2))) ∴ f(x)=2x^2 −5x+3 f(2)−f(−2)=1−21= determinant (((−20)))$
	$f (x) = a (x - 1) (x - 3) + g (x)$ ${ax}^{2} - 5 x + 3 = {ax}^{2} - 4 ax + 3 a + 3 x - 3$ ${ax}^{2} - 5 x + 3 = {ax}^{2} - (4 a - 3) x + 3 (a - 1)$ ${\begin{cases} 5 = 4 a - 3 \\ 1 = a - 1 \end{cases} \Rightarrow \begin{matrix} a = 2 \end{matrix}$ $∴ f (x) = {2 x}^{2} - 5 x + 3$ $f (2) - f (- 2) = 1 - 21 = \begin{matrix} - 20 \end{matrix}$
	Commented by dimentri last updated on 16/Jul/23

	$\underset{―}{\underset{⏟}{x}}$
	Answered by Rasheed.Sindhi last updated on 16/Jul/23
	$f(1)=g(1)=h a(1)^2 −5(1)+3=3(1)−3=h a−2=0 a=2 f(x)=2x^2 −5x+3 f(2)−f(−2) ={2(2)^2 −5(2)+3}−{2(−2)^2 −5(−2)+3} =1−21=−20 •“(3,t) is intersection point”is unnecessary.One common point is sufficient here.$
	$f (1) = g (1) = h$ $a {(1)}^{2} - 5 (1) + 3 = 3 (1) - 3 = h$ $a - 2 = 0$ $a = 2$ $f (x) = 2 x^{2} - 5 x + 3$ $f (2) - f (- 2)$ $= {2 {(2)}^{2} - 5 (2) + 3} - {2 {(- 2)}^{2} - 5 (- 2) + 3}$ $= 1 - 21 = - 20$ $∙ ‘ ‘ (3, t) i s i n t e r s e c t i o n {p o i n t}^{″} i s$ $u n n e c e s s a r y . O n e c o m m o n p o i n t$ $i s s u f f i c i e n t h e r e .$
	Commented by dimentri last updated on 16/Jul/23


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