The problem asks you to evaluate the function $f(x) = x^2 + 3x + 7$ for three different values: $f(-3)$, $f(x+1)$, and $f(-x)$. Let's go through each one step by step:

a. $f(-3)$

Substitute $x = -3$ into the function:

$f(-3) = (-3)^2 + 3(-3) + 7 = 9 - 9 + 7 = 7$

So, $f(-3) = 7$.

b. $f(x+1)$

Substitute $x+1$ into the function:

$f(x+1) = (x+1)^2 + 3(x+1) + 7$

Expanding the squares and simplifying:

$(x+1)^2 = x^2 + 2x + 1$ $3(x+1) = 3x + 3$

Now substitute into the function:

$f(x+1) = (x^2 + 2x + 1) + (3x + 3) + 7 = x^2 + 5x + 11$

So, $f(x+1) = x^2 + 5x + 11$.

c. $f(-x)$

Substitute $-x$ into the function:

$f(-x) = (-x)^2 + 3(-x) + 7$

Simplify:

$(-x)^2 = x^2$ $3(-x) = -3x$

Now substitute into the function:

$f(-x) = x^2 - 3x + 7$

So, $f(-x) = x^2 - 3x + 7$.

Final answers:

• $f(-3) = 7$
• $f(x+1) = x^2 + 5x + 11$
• $f(-x) = x^2 - 3x + 7$

Would you like further details on any of these steps?

Here are 5 follow-up questions to deepen your understanding:

1. How would you evaluate $f(x)$ for another arbitrary value, such as $f(2)$?
2. What happens to the function $f(x)$ if you substitute $x = 0$?
3. How would the result change for $f(x-1)$?
4. Can you derive the vertex form of the quadratic function $f(x) = x^2 + 3x + 7$?
5. How would you find the roots of $f(x) = 0$?

Tip: Always simplify expressions step by step to avoid mistakes in expanding or factoring terms.