The problem provides the graph of a quadratic function $f(x) = x^2 + kx + p$. From the graph, we know that the x-intercepts are at $x=0$ and $x=4$, and the y-coordinate of the vertex is $-8$. We need to find the values of $k$ and $p$, and then find the coordinates of point $A$, which is the x-intercept other than $x=0$.

1. Problem Description

The problem provides the graph of a quadratic function $f(x) = x^2 + kx + p$. From the graph, we know that the x-intercepts are at $x=0$ and $x=4$, and the y-coordinate of the vertex is $-8$. We need to find the values of $k$ and $p$, and then find the coordinates of point $A$, which is the x-intercept other than $x=0$.

2. Solution Steps

* Step 1: Find the value of $p$.

The y-intercept of the function $f(x) = x^2 + kx + p$ is the value of $f(0)$. From the given graph, we see that the parabola passes through the origin $(0,0)$, meaning $f(0)=0$. Therefore, $p=0$.

$f(0) = 0^2 + k(0) + p = 0$, so $p=0$.

Our function becomes $f(x) = x^2 + kx$.

* Step 2: Find the value of $k$.

The x-intercepts are $0$ and $4$. Since $f(4)=0$, we can substitute $x=4$ into the equation to solve for $k$.

$f(4) = 4^2 + k(4) = 0$

$16 + 4k = 0$

$4k = -16$

$k = -4$.

Thus, the function is $f(x) = x^2 - 4x$.

* Step 3: Verify the y-coordinate of the vertex.

The x-coordinate of the vertex of a quadratic $ax^2 + bx + c$ is $x = -\frac{b}{2a}$. For $f(x) = x^2 - 4x$, the x-coordinate of the vertex is $x = -\frac{-4}{2(1)} = 2$.

Then, $f(2) = (2)^2 - 4(2) = 4 - 8 = -8$. This confirms the given information about the vertex's y-coordinate.

* Step 4: Find the coordinates of point A.

Point $A$ is the x-intercept of the parabola other than $x=4$. We know that the parabola crosses the x-axis when $f(x) = 0$.

$x^2 - 4x = 0$

$x(x - 4) = 0$

This means $x=0$ or $x=4$. However, the graph indicates $A$ is the other x-intercept.

Since one x-intercept is at $x=0$ and the other at $x=4$ in the given image of the graph, this means that point $A$ must coincide with the intersection with the x-axis at x=

0. Therefore, the coordinates of A are (0,0) in this case. However, from the diagram A is clearly to the left of the y-axis.

If we assume instead that the x-intercepts are at x=0 and at some value that is NOT 4, then we can still continue. From the information given, the vertex of the parabola is at $x = 2$. The distance from the x-coordinate of the vertex to 0 is

2. Then, due to symmetry, the x-coordinate of the second x-intercept must be $2+2=4$. But from the diagram point A is to the left of the origin.

If we assume that x=4 is where the parabola cuts the x-axis furthest to the right, and that the second intersection is actually at the origin, then we can find where it cuts the graph again. This can happen if we say instead $f(x) = -(x^2 + kx + p)$ then this would be upside down, which would mean $p=8$. However, point (4,0) should satisfy it.

Given the information in the prompt, it seems very likely that the x intercepts are 0 and

4. So in this case, it must be that point A's x coordinate is close to 0, i.e.

0. Then $A=(0,0)$.

3. Final Answer

$k = -4$

$p = 0$

Coordinates of point $A$: $(0, 0)$