Given the function $f(x) = -\frac{1}{4}(x+2)(x-5)$, let's solve the problem step by step:

1. X-intercepts

The x-intercepts occur where $f(x) = 0$. This happens when either factor of the equation is zero:

$-\frac{1}{4}(x+2)(x-5) = 0$

For $(x+2) = 0$, $x = -2$,
For $(x-5) = 0$, $x = 5$.

Thus, the x-intercepts are: $x = -2 \quad \text{and} \quad x = 5$

2. Y-intercept

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function:

$f(0) = -\frac{1}{4}(0+2)(0-5) = -\frac{1}{4}(2)(-5) = -\frac{1}{4}(-10) = \frac{10}{4} = 2.5$

Thus, the y-intercept is $y = 2.5$.

3. Axis of Symmetry

The axis of symmetry of a quadratic function is the vertical line that passes through the midpoint of the x-intercepts. The x-intercepts are $x = -2$ and $x = 5$. The midpoint is:

$x = \frac{-2 + 5}{2} = \frac{3}{2} = 1.5$

Thus, the equation of the axis of symmetry is: $x = 1.5$

4. Graph

The function is a downward-opening parabola (because of the negative coefficient), with x-intercepts at $x = -2$ and $x = 5$, and the vertex along the axis of symmetry $x = 1.5$. The y-intercept is at $y = 2.5$.

If you'd like, I can generate a detailed graph for you based on this function!

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Do you want me to graph it for you? Feel free to ask if you need more explanation. Here are 5 related questions to help expand your understanding:

1. How do you find the vertex of a parabola from the factored form?
2. What does the sign of the leading coefficient tell you about the graph of the parabola?
3. How do you derive the formula for the axis of symmetry from the standard quadratic form?
4. What are the properties of a parabola that has two distinct x-intercepts?
5. How does completing the square help in finding the vertex?

Tip: The axis of symmetry formula can be quickly found as $x = \frac{-b}{2a}$ for a quadratic equation in standard form $ax^2 + bx + c$.