Answer: \(x\)-intercept \((-\frac12,0)\), \(y\)-intercept \((0,5\ln3)\), asymptote \(x=-\frac34\).

The graph is

\(y=5\ln(4x+3).\)

The logarithm is defined when

\(4x+3\gt0.\)

So

\(x\gt-\dfrac34.\)

Therefore the vertical asymptote is

\(x=-\dfrac34.\)

To find the \(x\)-intercept, put \(y=0\):

\(5\ln(4x+3)=0.\)

So

\(\ln(4x+3)=0.\)

This means

\(4x+3=1,\)

and therefore

\(x=-\dfrac12.\)

Thus the \(x\)-intercept is

\(\left(-\dfrac12,0\right).\)

To find the \(y\)-intercept, put \(x=0\):

\(y=5\ln3.\)

Thus the \(y\)-intercept is

\((0,5\ln3).\)

For the sketch, the graph is increasing for \(x\gt-\dfrac34\), tends to \(-\infty\) as \(x\) approaches \(-\dfrac34\) from the right, and increases slowly as \(x\) becomes large.