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I can see the image of math problems you've uploaded. Let's begin solving the questions provided one by one.

### Problem 14:
\[ f(x) = -x^3 + 2x^2 - 1 \quad \text{and} \quad f(-3) \]
Substitute \( x = -3 \) into the function:
\[
f(-3) = -(-3)^3 + 2(-3)^2 - 1
\]
\[
= -( -27 ) + 2(9) - 1 = 27 + 18 - 1 = 44
\]

Thus, \( f(-3) = 44 \).

### Problem 15:
\[ g(x) = |13 - x^2| \quad \text{and} \quad g(1) \]
Substitute \( x = 1 \) into the function:
\[
g(1) = |13 - 1^2| = |13 - 1| = |12| = 12
\]

Thus, \( g(1) = 12 \).

### Problem 16:
\[ p(x) = \frac{x^2 - 4}{2x + 1} \quad \text{and} \quad p(-8) \]
Substitute \( x = -8 \) into the function:
\[
p(-8) = \frac{(-8)^2 - 4}{2(-8) + 1} = \frac{64 - 4}{-16 + 1} = \frac{60}{-15} = -4
\]

Thus, \( p(-8) = -4 \).

### Problem 17:
\[ h(x) = 3 \cdot 9^x \quad \text{and} \quad h(-2) \]
Substitute \( x = -2 \) into the function:
\[
h(-2) = 3 \cdot 9^{-2} = 3 \cdot \frac{1}{9^2} = 3 \cdot \frac{1}{81} = \frac{3}{81} = \frac{1}{27}
\]

Thus, \( h(-2) = \frac{1}{27} \).

Would you like me to continue solving the rest of the problems, or provide further details on these solutions?

Evaluate each function for the given value for problems 14 to 17.

Explore how to evaluate polynomial, absolute value, rational, and exponential functions for specific inputs. Covering the methods to solve functions such as f(x) = -x^3 + 2x^2 - 1, g(x) = |13 - x^2|, p(x) = (x^2 - 4)/(2x + 1), and h(x) = 3 * 9^x. Ideal for high school students learning advanced algebra concepts.

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