Calculus! It's about differentiation and limit.?

Here's the question.

"f(x)=(x^3)+1. if x≤1
f(x)=a(x-2)^2 +b, if x>1

What values of the constants a+b make function f(x) differentiable at x=1? show that the limits of f(x)
as x approaches 1 from the right and from the left can equal each other but that f is still not
differentiable at x=1."

I know y1=y2, and y'1=y'2. so I found out a=-3/2 and b=7/2.
Therefore, a+b would be 2.
now I don't know what is next step that I should do. Do I need to draw a graph, or what?
Please help me if you know!

The statement is a little hard to understand---i.e. it's not too clear what the author is looking for.

We can notice that the continuity condition (on f, not f ') gives a + b = 2.

lim f(x) = 2 and
x->1-

lim f(x) = a + b.
x->1+

Hence a + b = 2.

Of course, there are lots of ways to make this true. The choice a = 0, b = 2 will do the trick. But as you've noticed, not just any old choice of a and b with a + b = 2 will give a differentiable function.

To make it differentiable, you have to satisfy the limits

lim 3x² = lim 2a(x - 2)
x->1- . . x->1+

which requires a = -3/2.

So the function is continuous for any a, b such that a + b = 2. It is differentiable if and only if a = -3/2 and b = 7/2.

a = -3/2, b = 7/2

you are done. Stop there.