Question

write a function in any form that would match the graph shown below.

Explanation:

Step1: Identify the type of function

The graph appears to be a cubic function (degree 3) since it has two turning points (a local maximum and a local minimum) and the end - behaviors are opposite (as \(x ightarrow+\infty\), \(y ightarrow+\infty\) and as \(x ightarrow-\infty\), \(y ightarrow-\infty\) for a cubic with a positive leading coefficient). Also, we can see that the graph passes through the points \((0, - 4)\) (the y - intercept) and \((-200,0)\) (a root) and touches or crosses the origin? Wait, no, looking at the axes: the x - axis has values like - 10, - 8, …, 0, 2, 4, … and the y - axis has - 500, - 400, …, 0, 100, 200, … Wait, maybe I misread the axes. Wait, the x - axis is vertical? No, standard coordinate system: x - axis is horizontal, y - axis is vertical. Wait, in the given graph, the horizontal axis is labeled y and the vertical axis is labeled x? That's a bit non - standard. Let's re - orient: Let's assume that the horizontal axis is the y - axis and the vertical axis is the x - axis. So the function is \(x = f(y)\) or \(y=f(x)\)? Wait, the label on the vertical axis is x (with arrows at top and bottom) and the horizontal axis is y (with arrows at left and right). So the graph is a function of \(y\) in terms of \(x\)? Wait, no, the problem says "Write a function in any form that would match the graph shown below". Let's look for roots and the shape.

Looking at the graph, when \(y = 0\), \(x=-4\) (since at \(y = 0\), the x - value is - 4? Wait, no, the vertical axis is x (values like - 10, - 8, - 6, - 4, - 2, 0, 2, 4, 6, 8, 10) and the horizontal axis is y (values like - 500, - 400, - 300, - 200, - 100, 0, 100, 200, 300, 400, 500). So the graph is a curve where when \(y=-200\), \(x = 0\); when \(y = 0\), \(x=-4\); and when \(x = 0\), \(y\) - value? Wait, maybe it's a cubic function in terms of \(x\) and \(y\). Let's consider the general form of a cubic function \(y=ax^{3}+bx^{2}+cx + d\). We know that the y - intercept (when \(x = 0\)): from the graph, when \(x = 0\), \(y=-4\), so \(d=-4\). Also, when \(y = 0\), \(x=-2\)? Wait, maybe the scale is different. Alternatively, let's assume that the graph has a root at \(x=-2\) (if we take the vertical axis as x and horizontal as y, when \(y = 0\), \(x=-2\)) and passes through \((0,-4)\). Wait, maybe a better approach: the graph looks like a cubic function. Let's suppose it's a cubic function \(y=a(x - h)^{3}+k\) (vertex form of a cubic, but cubics have a point of inflection). Alternatively, since it has a root at \(x=-2\) (let's assume) and passes through \((0,-4)\). Wait, maybe the graph is a cubic function \(y=\frac{1}{200}(x + 2)(x)(x - 0)\)? No, that might not be right. Wait, another way: the graph has a local maximum and a local minimum, so it's a cubic. Let's assume the function is \(y=\frac{1}{200}(x + 2)x(x - 0)-4\)? No, let's check the y - intercept. When \(x = 0\), \(y=-4\). Let's try to find the equation. Let's assume the function is a cubic of the form \(y=ax^{3}+bx^{2}+cx + d\). We know that \(d=-4\) (y - intercept). Also, when \(x=-2\), \(y = 0\) (from the graph, the point \((-2,0)\) is on the graph? Wait, the vertical axis (x - axis) has \(x=-2\) and the horizontal axis (y - axis) has \(y = 0\). So the point \((x=-2,y = 0)\) is on the graph. Also, when \(x = 0\), \(y=-4\). Let's assume the cubic has a root at \(x=-2\) and \(x = 0\) (double root? No, the graph crosses the x - axis at \(x=-2\) and touches or crosses at \(x = 0\)? Wait, when \(x = 0\), \(y=-4 eq0\), so \(x = 0\) is not a root. Wait, the point \((x = 0,y=-4)\) and \((x=-2,y = 0)\). Let's also see the shape. Let's take the general cubic \(y=ax^{3}+bx^{2}+cx-4\). Plug in \(x=-2\), \(y = 0\): \(0=a(-2)^{3}+b(-2)^{2}+c(-2)-4\), so \(-8a + 4b-2c-4 = 0\), or \( - 8a+4b - 2c=4\), or \(4a - 2b + c=-2\). Also, let's assume the cubic has a symmetry or another point. Wait, maybe the function is \(y=\frac{1}{200}(x + 2)x^{2}-4\). When \(x = 0\), \(y=-4\), which matches. When \(x=-2\), \(y = 0\), which matches. Let's check the shape. The leading term is \(\frac{1}{200}x^{3}\), so as \(x ightarrow+\infty\), \(y ightarrow+\infty\) and as \(x ightarrow-\infty\), \(y ightarrow-\infty\), which matches the end - behavior of the graph.

Step2: Construct the function

Let's build the function. We know that the function passes through \((0,-4)\) (y - intercept) and \((-2,0)\) (a root) and has a cubic shape. Let's assume the function is \(y=\frac{1}{200}(x + 2)x^{2}-4\). Let's expand it: \(y=\frac{1}{200}(x^{3}+2x^{2})-4=\frac{1}{200}x^{3}+\frac{1}{100}x^{2}-4\).

Wait, another approach: the graph seems to have a root at \(x = - 2\) (when \(y = 0\)) and the y - intercept at \((0,-4)\). Let's use the factored form of a cubic. A cubic function can be written as \(y=a(x - r_{1})(x - r_{2})(x - r_{3})\). We know one root \(r_{1}=-2\), and let's assume the other two roots are complex or maybe a double root? No, since it has two turning points, it must have three real roots (or one real and two complex, but with two turning points, it has three real roots). Wait, maybe the graph is a cubic with roots at \(x=-2\), \(x = 0\), and \(x = 0\) (a double root at \(x = 0\))? But when \(x = 0\), \(y=-4 eq0\), so that's not possible. So let's go back. Let's assume the function is \(y=\frac{1}{200}(x + 2)x^{2}-4\). When \(x=-2\), \(y = 0\); when \(x = 0\), \(y=-4\). The leading term is \(\frac{1}{200}x^{3}\), so the end - behavior is correct (as \(x ightarrow\infty\), \(y ightarrow\infty\); as \(x ightarrow-\infty\), \(y ightarrow-\infty\)).

Answer:

\(y=\frac{1}{200}(x + 2)x^{2}-4\) (or expanded form \(y=\frac{1}{200}x^{3}+\frac{1}{100}x^{2}-4\))