To solve the problem, we will substitute the given values into the function $h(x) = 2x^2 - x + 3$.

1. Find $h(3)$: $h(3) = 2(3)^2 - 3 + 3 = 2(9) - 3 + 3 = 18 - 3 + 3 = 18$

2. Find $h(2-)$: If by $h(2-)$, you mean approaching $x = 2$ from the left side (left-hand limit), the function $h(x)$ is continuous and does not have a discontinuity at $x = 2$, so we can directly substitute $x = 2$: $h(2) = 2(2)^2 - 2 + 3 = 2(4) - 2 + 3 = 8 - 2 + 3 = 9$ Hence, $h(2-) = h(2) = 9$.

3. Find $h(x+1)$: Substituting $x+1$ into the function: $h(x+1) = 2(x+1)^2 - (x+1) + 3$ First, expand $(x+1)^2$: $(x+1)^2 = x^2 + 2x + 1$ Now, substitute back into the equation: $h(x+1) = 2(x^2 + 2x + 1) - (x+1) + 3 = 2x^2 + 4x + 2 - x - 1 + 3$ Simplify: $h(x+1) = 2x^2 + 3x + 4$

So, the results are:

• $h(3) = 18$
• $h(2-) = 9$
• $h(x+1) = 2x^2 + 3x + 4$

Would you like any further details on these calculations?

Relative questions:

1. What is the value of $h(4)$?
2. How would you find $h(0)$?
3. What does it mean for a function to have a left-hand limit like $h(2-)$?
4. Can you derive the general form of $h(x-1)$?
5. How does the quadratic nature of the function $h(x)$ affect its graph?

Tip:

For polynomial functions like $h(x)$, continuity means the left-hand and right-hand limits will always be the same at any point.