The point that must be on the graph of:
[tex]y=5g(x+1)[/tex]
is:
[tex](3,1)[/tex]
In this case, we have the following function:
[tex]g(x)[/tex]
This function contains the point:
[tex](4,\frac{1}{5})[/tex]
In other words:
[tex]For \ the \ point \ (4,\frac{1}{5}) \rightarrow x=4 \ and \ y=\frac{1}{5} \\ \\ So: \\ \\ y=g(x) \ and \ g(4)=\frac{1}{5} \\ \\ \\ For \ y=5g(x+1) \\ \\ We \ know \ g(4) \ So: \\ \\ x+1=4 \ \therefore x=3 \\ \\ Then: \\ \\ y=5g(3+1) \\ \\ y=5g(4) \\ \\ y=5(\frac{1}{5}) \\ \\ y=1[/tex]
So the point that must be on the graph of:
[tex]y=5g(x+1)[/tex]
is:
[tex](3,1)[/tex]
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