How do you solve a logarithmic functions of the form $f(x) = \log_b (x+c) +d$

Question

$f(x) = \log_b (x+c) +d$

$x-\text{intercept:} \space 2$
$y-\text{intercept:} \space 1$

contains the point $(-1,2)$
asymptotic$: x = -2$

behavior: decreasing.

This question was given to us without being taught on how to do it. What method should be used?

Answer 1

You probably mean finding the values of $ b,c, d $ .

x-intercept: 2 means $\log_b (2+c) + d = 0$

y-intercept: 1 means $1 = \log_b c + d $

contains the point $ (-1,2) $ means $ 2= \log_b (-1 +c) + d $

decreasing behavior means $ f'(x) \le 0 $

Can you go on?

Answer 2

$1=\log_b(c)-\log_b(2+c)=\log_b(\frac{c}{2+c})$
So $b^1=b=\frac{c}{2+c}$
$\log_b(x-c)\to\infty$ as $x \to c$ . And as $x\to-2$, $x\to+\infty$. Therefore $c=2$. $b$ and $d$ are easy to evaluate with $c=2$