calculus value of constant

Question

Find the value of constants a and b so that the given function is continuous everywhere.

f(x)={-ax^2+bx x<2 , 3 x=2, a+bx x>2}

Answer 1

Since each of the three pieces (-ax^2+bx, 3, and a+bx) are continuous no matter what the values of a and b are, the only thing necessary is to connect them up at the same point and that point is when x = 2.

So what we want is, that when we plug in x = 2, the values of all three pieces are the same. Since the value of the middle piece is given (3), we just need to make sure the values of the 1st and 3rd pieces are 3 at x = 2 (even though those pieces are not valid at x = 2).

So -a(2)^2+b(2) = 3 and a+b(2) = 3. Simplifying, -4a + 2b = 3 and a + 2b = 3. Solving (method of your choice), a = 0 and b = 3/2