(1,1) is a point on the curve, which tells us that we can plug x=1 and y=1 into the function:
1=4/3+a+b+2/3=2+a+b, So a+b=-1. If we find a then we can find b. b=-a-1.
At a turning-point the gradient is zero, so we need to differentiate the function:
dy/dx=4x2+2ax+b=0. We are told that (1,1) is a turning-point, so x=1 is a zero of this quadratic derivative. Therefore a factor is x-1. The other factor is q(x-p), where p and q are constants we need to find or relate to a and b. Therefore:
4x2+2ax+b=q(x-1)(x-p)=q(x2-(p+1)x+p)=qx2-q(p+1)x+pq=0.
Equating coefficients, we can see that q=4 because we have 4x2;
We also know that b=pq=4p (constant term);
And the x term tells us that 4(p+1)=-2a, that is, 4p+4=-2a, that is, b+4=-2a.
But b=-a-1 so, -a-1+4=-2a, a=-3, b=-a-1=3-1=2.
CHECK
y=4x3/3-3x2+2x+2/3, by putting in a=-3 and b=2, our proposed solution.
Let's see what happens if we plug in x=1:
y=4/3-3+2+2/3=1, which proves that (1,1) lies on the curve.
Now differentiate:
4x2-6x+2=0 at a turning-point. (Incidentally, 4x2-6x+2 factorises: 4(x-1)(x-½) or 2(x-1)(2x-1). So p=½.) We were told that x=1 at a turning-point, so plug in x=1:
4-6+2=0 is true so x=1 is indeed a turning-point.
This confirms the result a=-3 and b=2.