Given that g(x) = ax2 + bx − 3, g(2) = 4 and g(-2) = 6. Find the values of a and b.

Substitute $x=2$ into the function $g(x) = ax^2 + bx - 3$:

$$g(2) = a(2)^2 + b(2) - 3$$ $$g(2) = 4a + 2b - 3$$

Since $g(2) = 4$, we set the expression equal to 4:

$$4a + 2b - 3 = 4$$ $$4a + 2b = 7 \quad (\text{Equation 1})$$

Substitute $x=-2$ into the function $g(x) = ax^2 + bx - 3$:

$$g(-2) = a(-2)^2 + b(-2) - 3$$ $$g(-2) = 4a - 2b - 3$$

Since $g(-2) = 6$, we set the expression equal to 6:

$$4a - 2b - 3 = 6$$ $$4a - 2b = 9 \quad (\text{Equation 2})$$

We have the system:

1. $4a + 2b = 7$
2. $4a - 2b = 9$ Add Equation 1 and Equation 2 to eliminate $b$:

$$(4a + 2b) + (4a - 2b) = 7 + 9$$ $$8a = 16$$ $$a = \frac{16}{8}$$ $$a = 2$$

Using Equation 1: $4a + 2b = 7$
Substitute $a=2$:

$$4(2) + 2b = 7$$ $$8 + 2b = 7$$ $$2b = 7 - 8$$ $$2b = -1$$ $$b = -\frac{1}{2}$$

The values are $a=2$ and $b=-\frac{1}{2}$.