hello mister. i need help explaining how determine the range of function of rational functions like (i) f(x)=((ax^2 +bx+c)/(px^2 +qx+r)) (ii) f(x)=((ax^2 +bx+c)/(px+q))


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Question Number 79816 by jagoll last updated on 28/Jan/20

$hello mister .$ $i need help explaining how determine$ $the range of function$ $of rational functions$ $like (i) f (x) = \frac{{ax}^{2} + bx + c}{{px}^{2} + qx + r}$ $(ii) f (x) = \frac{{ax}^{2} + bx + c}{px + q}$
	Answered by MJS last updated on 28/Jan/20
	$y=((ax^2 +bx+c)/(px^2 +qx+r)) leads to x^2 +((qy−b)/(py−a))+((ry−c)/(py−a))=0 D=(((4pr−q^2 )y^2 −2(2ar−bq+2cp)y+4ac−b^2 )/(4(py−a)^2 )) D<0 ⇒ no real solution for x ⇒ no y exists the range then is R\[interval for y where D<0] same when p=0 which is your (ii)$
	$y = \frac{{a x}^{2} + b x + c}{{p x}^{2} + q x + r}$ $leads to$ $x^{2} + \frac{q y - b}{p y - a} + \frac{r y - c}{p y - a} = 0$ $D = \frac{(4 p r - q^{2}) y^{2} - 2 (2 a r - b q + 2 c p) y + 4 a c - b^{2}}{4 {(p y - a)}^{2}}$ $D < 0 \Rightarrow no real solution for x \Rightarrow no y exists$ $the range then is R ∖ [interval for y where D < 0]$ $same when p = 0 which is your (ii)$
	Commented by jagoll last updated on 28/Jan/20
	$if the function y=((2x^2 −2)/(3x^2 +3x−2)) ⇒3x^2 y+3xy−2y=2x^2 −2 (3y−2).x^2 +3y.x+2−2y=0 D= 9y^2 −4.(3y−2)(2−2y) D= 9y^2 −8(3y−2)(1−y) D=9y^2 −8(−3y^2 +5y−2) D=33y^2 +40y+16 >0 for ∀y∈R R_f = {y∣ y∈R } that right sir?$
	$if the function$ $y = \frac{{2 x}^{2} - 2}{{3 x}^{2} + 3 x - 2} \Rightarrow {3 x}^{2} y + 3 xy - 2 y = {2 x}^{2} - 2$ $(3 y - 2) . x^{2} + 3 y . x + 2 - 2 y = 0$ $D = {9 y}^{2} - 4 . (3 y - 2) (2 - 2 y)$ $D = {9 y}^{2} - 8 (3 y - 2) (1 - y)$ $D = {9 y}^{2} - 8 (- {3 y}^{2} + 5 y - 2)$ $D = {33 y}^{2} + 40 y + 16 > 0 for \forall y \in R$ $R_{f} = {y ∣ y \in R}$ $that right sir ?$
	Commented by jagoll last updated on 28/Jan/20

	$range function not related to$ $flat asymtot sir ?$
	Commented by jagoll last updated on 28/Jan/20

	$for example$ $y = \frac{{2 x}^{2} + 3 x - 2}{{3 x}^{2} - 5 x + 2}$ $R_{f} : y \neq \lim_{x \to \infty} \frac{{2 x}^{2} + 3 x - 2}{{3 x}^{2} - 5 x + 2}$ $R_{f} : y \neq \frac{2}{3} ?$
	Commented by MJS last updated on 28/Jan/20

	$y = \frac{2 x^{2} - 2}{3 x^{2} + 3 x - 2}$ $x^{2} + \frac{3 y}{3 y - 2} x - \frac{2 (y - 1)}{3 y - 2} = 0$ $[x^{2} + P x + Q = 0 \Rightarrow D = \frac{P^{2}}{4} - Q]$ $D = \frac{33 y^{2} - 40 y + 16}{4 {(3 y - 2)}^{2}} > 0 \forall y \in R$ $\Rightarrow range is R$
	Commented by MJS last updated on 28/Jan/20
	$y=((2x^2 +3x−2)/(3x^2 −5x+2)) x^2 −((5y+3)/(3y−2))x+((2(y+1))/(3y−2))=0 D=((y^2 +22y+25)/(4(3y−2)^2 )) D<0 for −11−4(√6)<y<−11+4(√6) ⇒ range is R\]−11−4(√6); −11+4(√6)[$
	$y = \frac{2 x^{2} + 3 x - 2}{3 x^{2} - 5 x + 2}$ $x^{2} - \frac{5 y + 3}{3 y - 2} x + \frac{2 (y + 1)}{3 y - 2} = 0$ $D = \frac{y^{2} + 22 y + 25}{4 {(3 y - 2)}^{2}}$ $D < 0 for - 11 - 4 \sqrt{6} < y < - 11 + 4 \sqrt{6}$ $\Rightarrow range is R ∖] - 11 - 4 \sqrt{6}; - 11 + 4 \sqrt{6} [$
	Commented by jagoll last updated on 28/Jan/20

	$sir how to get in line 2 ? still understand$
	Commented by jagoll last updated on 28/Jan/20

	${3 x}^{2} y + 3 xy - 2 y - {2 x}^{2} + 2 = 0$ $(3 y - 2) x^{2} + (3 y) x + 2 - 2 y = 0$ $x^{2} + \frac{3 y}{3 y - 2} x + \frac{2 (1 - y)}{3 y - 2} = 0$ $that right sir ?$
	Commented by MJS last updated on 28/Jan/20

	$y = \frac{2 x^{2} + 3 x - 2}{3 x^{2} - 5 x + 2}$ $(3 x^{2} - 5 x + 2) y = 2 x^{2} + 3 x - 2$ $(3 y - 2) x^{2} - (5 y + 3) x + 2 (y + 1) = 0$ $x^{2} - \frac{5 y + 3}{3 y - 2} x + \frac{2 (y + 1)}{3 y - 2} = 0$ ${you}^{'} ve made mistakes in line 1$ $p = - \frac{5 y + 3}{3 y - 2}; q = \frac{2 (y + 1)}{3 y - 2}$ $D = \frac{p^{2}}{4} - q = \frac{y^{2} + 22 y + 25}{4 {(3 y - 2)}^{2}}$
	Commented by MJS last updated on 29/Jan/20

	$look at qu . 79883$
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