I've solved functions where it has a linear dividend and divisor only before and this is pretty new to me. I tried solving this by using what I've learnt in quadratics and polynomials but I keep going into a circle. Can anyone teach me on how would you go on solving a function where both the dividend and divisor are quadratics? Thank you.
On
Let $y$ be a real. You want to solve : $f(x)=y$. This means : \begin{align} \frac{(2x-1)(x+4)}{(x-1)(x+2)} &=y\iff (2x-1)(x+4)=y(x-1)(x+2) \\ \iff 2x^2+7x-4=yx^2+yx-2 \\ \iff (2-y)x^2+(7-y)x-2=0 \end{align} If $y\ne 2$, discriminant is : $$\Delta=(7-y)^2+8(2-y)=y^2-22y+65=(y-11)^2-56=(y-11-\sqrt{56})(y-11+\sqrt{56})$$ So : - if $y\in\left]11-\sqrt{56},11+\sqrt{56}\right[$, $\Delta<0$ and the equation has no solution, meaning $y$ doesn't have a pre-image; - if $y=11\pm\sqrt{56}$, the equation has one solution $x=-\frac{7-y}{4-2y}$, - if $y\in\left]-\infty,11-\sqrt{56}\right[ \cup \left]11+\sqrt{56},+\infty\right[$, the equation has two solutions : $$x=\frac{-(7-y)\pm \sqrt{y^2-22y+65}}{2(2-y)}$$ if $y=2$, the equation reduces to $5x-2=0$ so $x=\frac25$.
Hint
$f(x)=\frac{2x^2+7x-4}{x^2+x-2}$
$=2+\frac{5x}{x^2+x-2}=y$
you solve for $x \in(-\infty,-2)$
$,(-2,1)$ and $(1,+\infty)$.