Determine the value of $a$ and $b$ such that $(ax+3)/(bx+5)$ has $f (-2)= 1$ and $f '(-1) = 1/4$? I tried using the quotient rule but I honestly have no clue how to do this!
Can someone try to help?
$f(x)=\dfrac{ax+3}{bx+5}$.
Determine $a$ and $b$ such that $f(-2)=1$ and $f'(-1)=\frac14$.

I’m assuming that despite the missing required parentheses, $f(x)$ is supposed to be 
$$f(x)=\frac{ax+3}{bx+5}\;.$$
You do indeed want the quotient rule to differentiate $f$:
$$\begin{align*}
f\,'(x)&=\frac{(bx+5)(ax+3)'-(ax+3)(bx+5)'}{(bx+5)^2}\\\\
&=\frac{a(bx+5)-b(ax+3)}{(bx+5)^2}\\\\
&=\frac{abx+5a-abx-3b}{(bx+5)^2}\\\\
&=\frac{5a-3b}{(bx+5)^2}\;.
\end{align*}$$
Now you want $$1=f(-2)=\frac{-2a+3}{-2b+5}=\frac{3-2a}{5-2b}\tag{1}$$ and $$\frac14=f\,'(-1)=\frac{5a-3b}{(-b+5)^2}=\frac{5a-3b}{(5-b)^2}\;.\tag{2}$$
Multiply $(1)$ through by $5-2b$ to get $5-2b=3-2a$, and multiply $(2)$ through by $4(5-b)^2$ to get $(5-b)^2=4(5a-3b)$. Each of these can easily be solved for $a$ in terms of $b$:
$$\begin{align*}
a&=b-1\;,\text{ and}\\
a&=\frac1{20}\left(b^2+2b+25\right)\;.
\end{align*}\tag{3}$$
Now equate the righthand sides of $(3)$ to get a quadratic in $b$; you can solve that, and then use $b$ to find $a$.
