Let's solve the problem step by step.

The given quadratic equation is:
$y = -x^2 + 5x + p$

Key observations:

1. The graph is a parabola that opens downward because the coefficient of $x^2$ is negative ($-1$).
2. The vertex of the parabola is above the $x$-axis.
3. The graph intersects the $y$-axis at a negative point ($y < 0$).

We need to find the range of values of $p$.

Step 1: Vertex of the parabola

The $x$-coordinate of the vertex of a parabola $y = ax^2 + bx + c$ is given by:
$x_v = -\frac{b}{2a}$
Here, $a = -1$ and $b = 5$. So:
$x_v = -\frac{5}{2(-1)} = \frac{5}{2}$

Substitute $x_v = \frac{5}{2}$ into the equation $y = -x^2 + 5x + p$ to find the $y$-coordinate of the vertex: $y_v = -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) + p$ Simplify: $y_v = -\frac{25}{4} + \frac{25}{2} + p$ Convert $\frac{25}{2}$ to a denominator of 4: $y_v = -\frac{25}{4} + \frac{50}{4} + p = \frac{25}{4} + p$

Since the vertex lies above the $x$-axis ($y_v > 0$): $\frac{25}{4} + p > 0$ Solve for $p$: $p > -\frac{25}{4}$

Step 2: Intersecting the $y$-axis

The graph cuts the $y$-axis when $x = 0$. Substituting $x = 0$ into $y = -x^2 + 5x + p$:
$y = p$ Since the graph intersects the $y$-axis at a negative point: $p < 0$

Step 3: Combine the conditions

From the vertex condition, $p > -\frac{25}{4}$.
From the $y$-axis condition, $p < 0$.

Thus, the range of $p$ is: $-\frac{25}{4} < p < 0$

Final Answer:

The range of $p$ is: $-\frac{25}{4} < p < 0$

Would you like a detailed explanation for any part of this solution?

Here are some related questions to consider:

1. How do we determine the vertex of a parabola?
2. What does the condition $p < 0$ imply about the graph?
3. How do we know the graph opens downward?
4. How would the solution change if the vertex were below the $x$-axis?
5. How can we verify the graph satisfies the given conditions for specific values of $p$?

Tip: Always start analyzing quadratic equations by identifying the vertex and axis of symmetry.