Let's go through the math problem shown in the image.

We are given the function $y = f(x) = x^2 + 5$. There are four parts to the problem:

(a) Find the value of $y$ when $x = 0$:

Substituting $x = 0$ into the equation:

$y = 0^2 + 5 = 5$

So, the value of $y$ when $x = 0$ is $5$.

(b) What is $f(4)$?

Substituting $x = 4$ into the equation:

$f(4) = 4^2 + 5 = 16 + 5 = 21$

So, $f(4) = 21$.

(c) What values of $x$ give $y$ a value of $9$?

We are solving for $x$ when $y = 9$. So, we set $x^2 + 5 = 9$ and solve for $x$:

$x^2 + 5 = 9$ $x^2 = 9 - 5 = 4$ $x = \pm \sqrt{4} = \pm 2$

So, the values of $x$ are $-2$ and $2$.

(d) Are there any values of $x$ that give $y$ a value of $1$?

We are solving for $x$ when $y = 1$. Set $x^2 + 5 = 1$:

$x^2 + 5 = 1$ $x^2 = 1 - 5 = -4$

Since $x^2 = -4$ has no real solutions (because the square of a real number cannot be negative), there are no real values of $x$ that give $y = 1$.

It seems like your answer to part (c) is correct but maybe incomplete in some detail on the platform, so double-check if the formatting or additional explanations are needed.

Would you like more details or have any questions about this?

Here are 5 related questions you might find useful:

1. What are the roots of the equation $x^2 + 3x + 2 = 0$?
2. Can you solve $x^2 - 16 = 0$ for $x$?
3. How do you graph $y = x^2 + 5$ and what does the graph look like?
4. What happens to the graph of $y = x^2 + c$ as $c$ increases or decreases?
5. What are imaginary solutions, and how do they apply to equations like $x^2 + 4 = 0$?

Tip: When solving quadratic equations, check if you can factor or use the square root method before trying other techniques like the quadratic formula.