Answer: \(x\)-intercept \((1,0)\); tangent \(y-\ln5=\frac45(x-2)\).

The key point is to differentiate first, then substitute the required value or solve the resulting equation.

(a) The graph of

\(y=\ln(4x-3)\)

has domain

\(4x-3\gt 0,\)

so

\(x\gt \frac34.\)

Thus it has a vertical asymptote at

\(x=\frac34.\)

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

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

Hence

\(4x-3=1,\)

so

\(x=1.\)

The \(x\)-intercept is therefore \((1,0)\).

(b) Differentiate:

\(\frac{dy}{dx}=\frac4{4x-3}.\)

At \(x=2\),

\(y=\ln(8-3)=\ln5\)

and

\(\frac{dy}{dx}=\frac45.\)

So the tangent is

\(y-\ln5=\frac45(x-2).\)