Solve for a b and c…
f(x)=
 { ax2 + bx
bx + c		 x<2 x≥2
if it satisfies:
1>>> f(x) is continuous
2>>> f(3) = 2
3>>> limx→-1 f(x) = -3
Since we have three unknowns, to answer this, we need three equations. We can derive those by using the three (3) satisfaction criteria.

To satisfy the first Criterion f(x) is continuous, the Left Hand Sum and the Right Hand Sum should be equal.

ax2 + bx = bx + c

ax2 + bx = bx + c

ax2 = c

Substitute x=2.

a(2)2 = c

4a =c ← Eq (1)

Expressing the second criterion, f(3) =2, in equation form we have:

b(3) + c = 2

3b + c = 2 ← Eq (2)

We used bx + c because x=3 falls under x≥2 range.

For the third criterion, limx→-1 f(x) = -3, simply evaluate the limit in the equation where x=-1 falls under.

a(-1)2 + b(-1) = -3

a – b = -3 ← Eq (3)

----

We now have three equations so we can solve simultaneously:

Substitute Eq(1) to Eq(2)

4a + 3b = 2

Add thrice Eq(3) from Eq (2)

4a + 3b = 2
3a - 3b = -9

7a = -7
a = -1

a - b = -3
(-1) - b = -3
b = 3 - 1
b = 2

4a = c
4(-1) = c
c = -4

a = -1; b = 2; c = -4.


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