Thread: Continuity of a Piecewise Function

I need this answered as soon as possible please

y={(x^2-4)/(x-2) if x is less than 2
{ax^2-bx+3 if x is greater than and equal to 2 and less than 3
{2x-a+b if x is greater than and equal to 3

find a and b

I have done some work and found that a=(-3/2) and b=(-7/2) but this does not work for the continuity at the point x=3
please help

Thank you
Dylan Renke

is the (a + b) in the third equation in brackets

the answer for the equation is

a = 0.5
b = 0.5

plug it in and all the piecewise functions will work because all the equations will continue at x = 2

i tried that, but then when approaches 2 from the left, it equals 4, and from the right it equals 2

the a and b are not in brackets

(x^2 - 4) / (x - 2) = (x+2)(x-2) / (x-2) = (x +2)
x= 2 then (2 + 2) = 4

ax^2 - bx + 3
a =0.5
b = 0.5
then (0.5 * 2^2 - 0.5 * 2) +3
1 + 3 = 4

2x - a + b
2 * 2 - 0.5 + 0.5
=4

I am just too tired, i have tried everything and just keep messing up on stupid things, the answer is 0.5 and 0.5, thank you for the help (showing work took me like an hour cause of stupid mistakes)