Write the equation, in standard form, that represents function f shown in the graph.

The graph is a parabola with roots at x = -4 and x = 0. The vertex is at (-2, -1). The standard form of a quadratic equation is $f(x) = a(x-h)^2 + k$. Using the vertex and one other point, we find $a=1$. Thus, $f(x) = (x+2)^2 - 1$, which simplifies to $f(x) = x^2 + 4x + 4 - 1 = x^2 + 4x + 3$. Alternatively, using the roots, $f(x) = a(x-r_1)(x-r_2) = a(x+4)(x)$. Using the vertex (-2,-1), $-1 = a(-2+4)(-2) = a(2)(-2) = -4a$, so $a = 1/4$. This gives $f(x) = \frac{1}{4}(x+4)(x) = \frac{1}{4}(x^2+4x) = \frac{1}{4}x^2 + x$. Let's re-examine the vertex. The vertex appears to be at (-2, -1). The roots are at (-4, 0) and (0, 0). This means the vertex x-coordinate is the average of the roots: $(-4+0)/2 = -2$. The y-coordinate of the vertex is given as -1. So the vertex is indeed (-2, -1). The standard form is $f(x) = a(x-h)^2 + k$. Substituting the vertex (-2, -1): $f(x) = a(x - (-2))^2 + (-1) = a(x+2)^2 - 1$. Now use a point, e.g., (0, 0): $0 = a(0+2)^2 - 1 = 4a - 1$. So $4a = 1$, $a = 1/4$. The equation is $f(x) = \frac{1}{4}(x+2)^2 - 1$. Expanding this: $f(x) = \frac{1}{4}(x^2 + 4x + 4) - 1 = \frac{1}{4}x^2 + x + 1 - 1 = \frac{1}{4}x^2 + x$. Let's check the point (-4, 0): $f(-4) = \frac{1}{4}(-4)^2 + (-4) = \frac{1}{4}(16) - 4 = 4 - 4 = 0$. This is correct. Let's check the point (-3, -1). This point is labeled on the graph. $f(-3) = \frac{1}{4}(-3)^2 + (-3) = \frac{1}{4}(9) - 3 = \frac{9}{4} - \frac{12}{4} = -\frac{3}{4}$. This does not match (-3, -1). There seems to be a discrepancy in the provided points or the graph. Let's assume the roots (-4,0) and (0,0) and the vertex (-2,-1) are correct. The equation derived is $f(x) = \frac{1}{4}x^2 + x$. The standard form is $f(x) = ax^2 + bx + c$. The problem asks for standard form, which usually means $ax^2+bx+c$. The vertex form is $a(x-h)^2+k$. If standard form means $ax^2+bx+c$, then $f(x) = \frac{1}{4}x^2 + x$. Let's re-examine the points. (-4,0), (0,0), (-3,-1). If (-3,-1) is a point on the parabola, and (-4,0) and (0,0) are roots, then the vertex x-coordinate is -2. The y-coordinate of the vertex would be $f(-2) = \frac{1}{4}(-2)^2 + (-2) = \frac{1}{4}(4) - 2 = 1 - 2 = -1$. So the vertex is (-2,-1). The point (-3,-1) is NOT on this parabola. Let's assume the points (-4,0), (0,0) and (-3,-1) are correct. The vertex is at x = -2. The y-coordinate of the vertex is not -1. Let's use the points (-4,0), (0,0) and (-3,-1). $f(x) = ax^2+bx+c$. Since (0,0) is a root, c=0. So $f(x) = ax^2+bx$. Using (-4,0): $0 = a(-4)^2 + b(-4) = 16a - 4b$. So $4a = b$. Using (-3,-1): $-1 = a(-3)^2 + b(-3) = 9a - 3b$. Substitute $b=4a$: $-1 = 9a - 3(4a) = 9a - 12a = -3a$. So $a = 1/3$. Then $b = 4a = 4/3$. The equation is $f(x) = \frac{1}{3}x^2 + \frac{4}{3}x$. Let's check the vertex for this equation. Vertex x-coordinate = $-b/(2a) = -(4/3)/(2/3) = -2$. Vertex y-coordinate = $f(-2) = \frac{1}{3}(-2)^2 + \frac{4}{3}(-2) = \frac{4}{3} - \frac{8}{3} = -\frac{4}{3}$. So the vertex is (-2, -4/3). This does not match the graph which shows the vertex at (-2, -1).

Let's assume the vertex is (-2,-1) and the roots are (-4,0) and (0,0). This is contradictory as shown above.
Let's assume the vertex is (-2,-1) and the point (-4,0) is on the graph. Then $f(x) = a(x+2)^2 - 1$. $0 = a(-4+2)^2 - 1 = a(-2)^2 - 1 = 4a - 1$. So $a = 1/4$. $f(x) = \frac{1}{4}(x+2)^2 - 1 = \frac{1}{4}(x^2+4x+4) - 1 = \frac{1}{4}x^2 + x + 1 - 1 = \frac{1}{4}x^2 + x$.
Let's check the point (0,0). $f(0) = \frac{1}{4}(0)^2 + 0 = 0$. This works.
Let's check the point (-3,-1). $f(-3) = \frac{1}{4}(-3)^2 + (-3) = \frac{9}{4} - 3 = \frac{9-12}{4} = -\frac{3}{4}$. This does not match.

Let's assume the points (-4,0), (0,0) and (-3,-1) are correct. As calculated before, $f(x) = \frac{1}{3}x^2 + \frac{4}{3}x$. The vertex is (-2, -4/3). The graph shows the vertex at (-2, -1).

Let's assume the vertex is (-2,-1) and the point (-3,-1) is on the graph. This means the parabola is symmetric about x=-2, and the point (-3,-1) is one unit to the left of the axis of symmetry. The point one unit to the right of the axis of symmetry would be (-1, -1). Let's check if (-1,-1) is on the graph. It appears to be.
So, vertex is (-2,-1). $f(x) = a(x+2)^2 - 1$. Use point (-3,-1): $-1 = a(-3+2)^2 - 1 = a(-1)^2 - 1 = a - 1$. So $a=0$. This is not a parabola.

Let's assume the points (-4,0) and (0,0) are roots. The vertex x-coordinate is -2. The y-coordinate of the vertex is given as -1. So vertex is (-2,-1).
The equation in vertex form is $f(x) = a(x-h)^2 + k$.
$f(x) = a(x-(-2))^2 + (-1) = a(x+2)^2 - 1$.
Using the root (0,0): $0 = a(0+2)^2 - 1 = 4a - 1$. So $a = 1/4$.
The equation is $f(x) = \frac{1}{4}(x+2)^2 - 1$.
Expanding this to standard form $ax^2+bx+c$:
$f(x) = \frac{1}{4}(x^2 + 4x + 4) - 1 = \frac{1}{4}x^2 + x + 1 - 1 = \frac{1}{4}x^2 + x$.
Let's check the point (-3,-1).
$f(-3) = \frac{1}{4}(-3)^2 + (-3) = \frac{9}{4} - 3 = \frac{9-12}{4} = -\frac{3}{4}$.
The point (-3,-1) is marked on the graph. This point is not on the parabola defined by the roots (-4,0), (0,0) and vertex (-2,-1).

Let's assume the points (-4,0) and (-2,0) are roots. Then the vertex x-coordinate is $(-4-2)/2 = -3$. The y-coordinate of the vertex is given as -1. So vertex is (-3,-1).
The equation in vertex form is $f(x) = a(x-h)^2 + k$.
$f(x) = a(x-(-3))^2 + (-1) = a(x+3)^2 - 1$.
Using the root (0,0) from the graph, which is not a root. Let's use the root (-2,0).
$0 = a(-2+3)^2 - 1 = a(1)^2 - 1 = a - 1$. So $a = 1$.
The equation is $f(x) = (x+3)^2 - 1$.
Expanding this: $f(x) = x^2 + 6x + 9 - 1 = x^2 + 6x + 8$.
Let's check the roots: $x^2 + 6x + 8 = 0 \implies (x+2)(x+4) = 0$. Roots are -2 and -4. This matches the points (-4,0) and (-2,0).
Let's check the vertex: x-coordinate is -3. $f(-3) = (-3)^2 + 6(-3) + 8 = 9 - 18 + 8 = -1$. Vertex is (-3,-1). This matches the point (-3,-1).
So the equation is $f(x) = x^2 + 6x + 8$.

The problem asks for standard form, which is $ax^2+bx+c$.
The points labeled are (-4,0), (-2,0), and (-3,-1).
The graph shows a parabola opening upwards.
The roots are at x = -4 and x = -2.
The vertex is at x = (-4 + -2) / 2 = -3.
The y-coordinate of the vertex is given as -1. So the vertex is (-3, -1).
The equation of a parabola in vertex form is $f(x) = a(x-h)^2 + k$.
Substituting the vertex (-3, -1): $f(x) = a(x - (-3))^2 + (-1) = a(x+3)^2 - 1$.
To find 'a', we can use one of the roots, for example, (-2, 0).
$0 = a(-2+3)^2 - 1$
$0 = a(1)^2 - 1$
$0 = a - 1$
$a = 1$.
So the equation in vertex form is $f(x) = 1(x+3)^2 - 1$.
Now, expand this to standard form $f(x) = ax^2 + bx + c$.
$f(x) = (x^2 + 6x + 9) - 1$
$f(x) = x^2 + 6x + 8$.
Let's verify with the other root (-4, 0):
$f(-4) = (-4)^2 + 6(-4) + 8 = 16 - 24 + 8 = 0$. This is correct.
The standard form of the equation is $f(x) = x^2 + 6x + 8$.