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 $f'(-1)=\frac14$.

Write the function $f$ has $g/h$ with $g(x) = ax + 3$ and $h(x) = bx + 5$. Then
$$g'(-1) = a \text{ and } h'(-1) = b$$
Hence by the quotient rule, we see that
$$f'(-1) = \frac{h(-1) g'(-1) - h'(-1) g(-1)}{h(-1)^2} = \frac{(5 - b)a - b(3 - a)}{(5 - b)^2} = \frac{5a - 3b}{(5 - b)^2}$$
Now using the fact that $4f'(-1) = 1$, we see that
$$20a - 12b = (5 - b)^2$$
On the other hand, using the fact that $f(-2) = 1$, we have another equation
$$3 - 2a = 5 - 2b$$
Use these equations to solve for $a$ and $b$.
