Let's address Question 4, which asks for all intercepts (x and y) for the function $f(x) = x^2 - 4$.

Y-intercept:

The y-intercept occurs when $x = 0$. Substituting into the function: $f(0) = 0^2 - 4 = -4$ Thus, the y-intercept is $(0, -4)$.

X-intercepts:

The x-intercepts occur when $f(x) = 0$. So, set the equation equal to zero: $x^2 - 4 = 0$ $x^2 = 4$ $x = \pm 2$ Thus, the x-intercepts are $(-2, 0)$ and $(2, 0)$.

Final Answer:

The intercepts are:

• X-intercepts: $(-2, 0)$ and $(2, 0)$
• Y-intercept: $(0, -4)$

So, the correct answer is:
$(0, -4), (-2, 0), (2, 0)$

Let me know if you'd like further details! Here are 5 additional questions that build upon this:

1. How do you find intercepts for any general quadratic function?
2. Can a parabola have more than two x-intercepts?
3. What does the vertex of this parabola represent?
4. How would you solve $f(x) = x^2 - 4$ using the quadratic formula?
5. What would happen to the intercepts if the equation were $f(x) = x^2 + 4$?

Tip: Remember that the x-intercepts of a function occur when the output (y-value) is zero, and the y-intercept occurs when the input (x-value) is zero.