Answer: \(y=\dfrac13(2x-5)^{3/2}+4x-20\).

Integrate the second derivative once:

\(\frac{dy}{dx} = \int (2x-5)^{-1/2}\,dx.\)

Since the derivative of \(2x-5\) is \(2\),

\(\int (2x-5)^{-1/2}\,dx = (2x-5)^{1/2}+c.\)

So

\(\frac{dy}{dx}=(2x-5)^{1/2}+c.\)

The gradient is \(6\) when \(x=\dfrac92\). Substitute this:

\(6=\left(2\cdot\frac92-5\right)^{1/2}+c.\)

Now

\(2\cdot\frac92-5=4,\)

so

\(6=2+c.\)

Hence \(c=4\), and

\(\frac{dy}{dx}=(2x-5)^{1/2}+4.\)

Integrate again:

\(y=\int\left((2x-5)^{1/2}+4\right)\,dx.\)

This gives

\(y=\frac13(2x-5)^{3/2}+4x+d.\)

Use the point \(\left(\dfrac92,\dfrac23\right)\):

\(\frac23 = \frac13(4)^{3/2}+4\left(\frac92\right)+d.\)

Since \(4^{3/2}=8\),

\(\frac23=\frac83+18+d.\)

Therefore

\(d=-20.\)

Hence the equation of the curve is

\(\boxed{y=\frac13(2x-5)^{3/2}+4x-20}.\)