Can you help me with Calculus

Hey I am kind of struggling in Calculus at the moment. My class just started doing stuff with limits and I'm quite confused.

Specifically this problem:

Find the values of a and b that make ƒ continuous everywhere.

ƒ(x) =
{ (x^2 - 4)/(x - 2) while x < 2
{ ax^2 - bx + 3 while 2 < x < 3
{ 2x - a + b while x >= 3

Any help would be appreciated (i.e. someone kind enough to work out the problem) or even a link to a site that helps explain a similar problem.

Thanks

..........{ (x² - 4)/(x - 2), for x < 2
f(x) = { ax² + bx + 3, for 2 < x < 3
..........{ 2x - a + b, for x ≥ 3

(You seem to have left out a less/greater than or equal to symbol between the first 2 functions. I'll assume it's on the second one)

For f to be continuous at x = 2, the limits from the left and right must match as x approaches 2.

lim x⟶2⁺ f(x) = lim x⟶2⁻⁻ f(x)
The appropriate f(x) is determined by the piecewise portion defined above. Since the right hand limit implies the x value is greater than 2 and approaching 2, you use the "second" function. For the left hand limit, you use the "first" function.

lim x⟶2⁺ (ax² + bx + 3) = a(2)² + b(2) + 3
= 4a + 2b + 3

lim x⟶2⁻⁻ (x² - 4)/(x - 2)
lim x⟶2⁻⁻ (x + 2)(x - 2)/(x - 2)
lim x⟶2⁻⁻ (x + 2) = 2 + 2 = 4

So 4a + 2b + 3 = 4, or
4a + 2b = 1
- - - - - - - - - - - - - - -
For f to be continuous at x = 3, the following must be true:

lim x⟶3⁺ f(x) = lim x⟶3⁻⁻ f(x)

lim x⟶3⁺ (2x - a + b) = 2(3) - a + b
= 6 - a + b

lim x⟶3⁻⁻ (ax² + bx + 3) = a(3)² + b(3) + 3
= 9a + 3b + 3

So 6 - a + b = 9a + 3b + 3, or
10a + 2b = 3
- - - - - - - - - - - - - - -

Now you have a system of equations:
(i) 4a + 2b = 1
(ii) 10a + 2b = 3

Solve any way you like. I prefer elimination:
(i) -4a - 2b = -1
(ii) 10a + 2b = 3
= = = = = = = = =
6a = 2
a = 1/3

4(1/3) + 2b = 1
2b = -1/3
b = -1/6