The graph of the logarithmic function that passes through (-2,-2) and (1,0) is [tex]\log_2(x+3)-2[/tex].
Suppose that the function is
[tex]f(x)=\log_2(x+c)+d[/tex]
Then there are two observations that we must consider from the figure
- The graph must pass through (-2,-2) and (1,0)
So,
[tex]-2=\log_2(-2+c)+d[/tex]
[tex]\implies -2+c=e^{-2-d} ...~...~(1)[/tex]
And,
[tex]0=\log_2(1+c)+d\\\implies 1+c=e^{-d} ~...~...~(2)[/tex]
Observe the figure given below we see that the graph of
[tex]\log_2(x)[/tex] passes through (1,0) but does not pass through (-2,-2) and the graph is to the right of (-2,-2) hence, we need the graph to move left.
When we increase c the graph [tex]\log_2(x+c)+d[/tex] moves left and
When we decrease d then the graph [tex]\log_2(x+c)+d[/tex] moves down.
That makes the graph required to be [tex]\log_2(x+3)-2[/tex].
Hence, the graph of the logarithmic function that passes through (-2,-2) and (1,0) is [tex]\log_2(x+3)-2[/tex].
Learn more about logarithms here-
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