Answer:
[tex]y =log_e(x+3)[/tex]
Step-by-step explanation:
It is given that the graph corresponds to a natural logarithmic function.
That means, the function [tex]y[/tex] has a natural log (Log with base [tex]e[/tex]) of some terms of x.
It is given that asymptote of given curve is at [tex]x= -3[/tex]. i.e. when we put value
[tex]x= -3[/tex], the function will have a value [tex]y \rightarrow \infty[/tex].
We know that natural log of 0 is not defined.
So, we can say the following:
[tex]log_e(x+a)[/tex] is not defined at [tex]x= -3[/tex]
[tex]\Rightarrow x+a =0\\\Rightarrow x = -a[/tex]
i.e. [tex]x =-a[/tex] is the point where [tex]y \rightarrow \infty[/tex]
a = 3
Hence, the function becomes:
[tex]y =log_e(x+3)[/tex]
Also, given that the graph crosses x axis at x = -2
When we put x = -2 in the function:
[tex]y =log_e(-2+3) = log_e(1) = 0[/tex]
And y axis at 1.
Put x = 0, we should get y = 1
[tex]y =log_e(0+3) = log_e(3) \approx 1[/tex]
So, the function is: [tex]y =log_e(x+3)[/tex]
