Question

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

Explanation:

Step1: Identify roots and multiplicity

The graph touches the x - axis at \(x = - 6\) (so multiplicity 2) and crosses at \(x=-1\) (wait, looking at the graph, it crosses at \(x = 0\)? Wait no, the graph crosses the x - axis at \(x=-1\)? Wait, no, let's re - examine. The graph touches the x - axis at \(x=-6\) (so a double root) and crosses at \(x = 0\)? Wait, the y - intercept is at \((0,-30)\). Wait, let's see the x - intercepts: one at \(x=-6\) (touching, so factor \((x + 6)^2\)) and one at \(x = 0\) (crossing, so factor \(x\))? Wait, no, the graph crosses the x - axis at \(x=-1\)? Wait, no, the graph goes through the origin? Wait, the graph passes through \((0,0)\)? Wait, the y - axis intersection: the point is \((0,-30)\)? Wait, no, the graph has a point at \((0,-30)\)? Wait, maybe I misread. Let's look again. The graph touches the x - axis at \(x=-6\) (so a root of multiplicity 2) and crosses at \(x = 0\)? Wait, no, the graph crosses the x - axis at \(x=-1\)? Wait, no, let's think about the general form of a polynomial. Since it's a cubic? Wait, the end - behaviors: as \(x\to-\infty\), \(y\to+\infty\) and as \(x\to+\infty\), \(y\to-\infty\), so the leading coefficient is negative and the degree is odd. Let's assume it's a cubic function. Let's say the roots are \(x=-6\) (multiplicity 2) and \(x = 0\) (multiplicity 1). So the function is \(y=a(x + 6)^2x\). Now, we can find \(a\) using the y - intercept. The y - intercept is when \(x = 0\), but wait, when \(x = 0\), \(y=a(0 + 6)^2(0)=0\), but the graph has a y - intercept at \((0,-30)\)? Wait, maybe the root at \(x = 0\) is not there. Wait, maybe the roots are \(x=-6\) (multiplicity 2) and \(x=-1\)? Wait, no, let's check the graph again. The graph touches the x - axis at \(x=-6\) (so \((x + 6)^2\)) and crosses at \(x=-1\)? Wait, no, the graph crosses the x - axis at \(x = 0\)? Wait, I think I made a mistake. Let's look at the graph: when \(x = 0\), \(y=-30\). Let's assume the function is \(y=a(x + 6)^2(x + 1)\)? Wait, no, when \(x = 0\), \(y=a(6)^2(1)=36a\). If \(y=-30\) when \(x = 0\), then \(36a=-30\), so \(a=-\frac{30}{36}=-\frac{5}{6}\). Wait, but let's check the end - behavior. If the function is \(y=-\frac{5}{6}x(x + 6)^2\). Let's expand it: \(y=-\frac{5}{6}x(x^{2}+12x + 36)=-\frac{5}{6}(x^{3}+12x^{2}+36x)=-\frac{5}{6}x^{3}-10x^{2}-30x\). Let's check the y - intercept: when \(x = 0\), \(y = 0\), but the graph has a y - intercept at \((0,-30)\)? Wait, no, maybe the root is at \(x = 0\) with multiplicity 1, and \(x=-6\) with multiplicity 2. Let's plug \(x = 0\) into \(y=a(x + 6)^2x\), we get \(y = 0\), but the graph has \(y=-30\) at \(x = 0\)? Wait, maybe I misread the y - intercept. Maybe the y - intercept is \((0,-30)\), so when \(x = 0\), \(y=-30\). Let's assume the function is \(y=a(x + 6)^2(x - k)\). Wait, no, let's try a different approach. The graph touches the x - axis at \(x=-6\) (so \((x + 6)^2\)) and crosses at \(x = 1\)? No, this is getting confusing. Wait, let's consider that the function is a cubic, so let's write it as \(y=a(x + 6)^2(x)\). Now, when \(x = 0\), \(y = 0\), but the graph has a point at \((0,-30)\)? Wait, maybe the graph is \(y=-\frac{5}{6}x(x + 6)^2\). Let's check the value at \(x = 0\): \(y = 0\), but the graph shows a point at \((0,-30)\). Wait, maybe the root is at \(x = 0\) with multiplicity 1, and \(x=-6\) with multiplicity 2, and we use the point \((0,-30)\) to find \(a\). Wait, if \(y=a(x + 6)^2x\), and when \(x = 0\), \(y = 0\), but the graph has \(y=-30\) at \(x = 0\), so maybe my initial assumption about the roots is wrong. Wait, maybe the roots are \(x=-6\) (multiplicity 2) and \(x = 1\)? No, let's look at the graph again. The graph touches the x - axis at \(x=-6\) (so \((x + 6)^2\)) and crosses at \(x=-1\). Let's assume the function is \(y=a(x + 6)^2(x + 1)\). Then, when \(x = 0\), \(y=a(6)^2(1)=36a\). If \(y=-30\) when \(x = 0\), then \(36a=-30\), so \(a=-\frac{30}{36}=-\frac{5}{6}\). So the function is \(y=-\frac{5}{6}(x + 6)^2(x + 1)\). Let's expand this: \(y=-\frac{5}{6}(x^{2}+12x + 36)(x + 1)=-\frac{5}{6}(x^{3}+13x^{2}+48x + 36)=-\frac{5}{6}x^{3}-\frac{65}{6}x^{2}-40x - 30\). Let's check the end - behavior: as \(x\to-\infty\), the leading term \(-\frac{5}{6}x^{3}\to+\infty\) (since \(x^{3}\to-\infty\) and multiplied by \(-\frac{5}{6}\) gives \(+\infty\)), and as \(x\to+\infty\), \(-\frac{5}{6}x^{3}\to-\infty\), which matches the graph. Also, at \(x=-6\), the function has a double root, so it touches the x - axis there, and at \(x=-1\), it crosses the x - axis. And at \(x = 0\), \(y=-30\), which matches the y - intercept. Alternatively, we can write it in factored form: \(y=-\frac{5}{6}x(x + 6)^2\) is wrong because at \(x = 0\) it gives \(y = 0\), but the correct y - intercept is \(-30\). Wait, maybe the root at \(x = 0\) is not there. Wait, maybe the graph is \(y=-\frac{5}{6}(x + 6)^2(x + 1)\) as above.

Step2: Determine the leading coefficient

We know that the function has a root at \(x=-6\) (multiplicity 2) and \(x=-1\) (multiplicity 1). So the factored form is \(y=a(x + 6)^2(x + 1)\). We use the point \((0,-30)\) to find \(a\). Substitute \(x = 0\) and \(y=-30\) into the equation:
\(-30=a(0 + 6)^2(0 + 1)\)
\(-30=a\times36\times1\)
\(a=\frac{-30}{36}=-\frac{5}{6}\)

Step3: Write the function

Substitute \(a =-\frac{5}{6}\) into the factored form:
\(y=-\frac{5}{6}(x + 6)^2(x + 1)\)
We can also expand it, but the factored form is acceptable.

Answer:

\(y =-\frac{5}{6}(x + 6)^2(x + 1)\) (or an equivalent form like \(y=-\frac{5}{6}x^{3}-\frac{65}{6}x^{2}-40x - 30\))