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 \(x=0\) and \(x = - 1\) (assuming the crossing near \(x=-1\) and \(x = 0\)). Wait, looking at the graph, the x - intercepts: it touches at \(x=-6\) (a double root) and crosses at \(x = 0\) and maybe \(x=-1\)? Wait, the y - intercept is at \((0,-30)\). Let's assume the function is a cubic? No, the end - behavior: 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. Wait, the graph has a touch point at \(x=-6\) (multiplicity 2) and a cross at \(x = 0\) and maybe \(x=-1\)? Wait, let's re - examine. The graph touches the x - axis at \(x=-6\) (so factor \((x + 6)^2\)) and crosses at \(x = 0\) (factor \(x\)) and maybe \(x=-1\)? Wait, when \(x = 0\), \(y=-30\). Let's assume the function is \(y=a(x + 6)^2x(x + 1)\). Wait, no, the end - behavior: degree 4? No, as \(x\to-\infty\), \(y\to+\infty\) and \(x\to+\infty\), \(y\to-\infty\) implies odd degree. Wait, maybe the graph has a root at \(x=-6\) (multiplicity 2), \(x = 0\) (multiplicity 1), and \(x=-1\) (multiplicity 1). So the function is \(y=a(x + 6)^2x(x + 1)\). Now, use the y - intercept \((0,-30)\). Plug \(x = 0\), \(y=-30\) into the function: \(-30=a(0 + 6)^2(0)(0 + 1)\), which is 0. That's wrong. So maybe the roots are \(x=-6\) (multiplicity 2) and \(x = 0\) (multiplicity 1), and another root? Wait, the graph crosses the x - axis at \(x = 0\) and touches at \(x=-6\). Wait, maybe it's a cubic function? No, the touch at \(x=-6\) (multiplicity 2) and cross at \(x = 0\) (multiplicity 1), so degree 3? But degree 3: as \(x\to-\infty\), if leading coefficient is negative, \(y\to+\infty\) and \(x\to+\infty\), \(y\to-\infty\), which matches. Wait, let's try \(y=a(x + 6)^2x\). Then, when \(x = 0\), \(y = 0\), but the y - intercept is \((0,-30)\). So that's not right. Wait, maybe the function is \(y=a(x + 6)^2(x)(x + 1)\) is degree 4, but end - behavior for degree 4: if leading coefficient is negative, as \(x\to\pm\infty\), \(y\to-\infty\), which doesn't match. Wait, I made a mistake in end - behavior. Let's look again: as \(x\to-\infty\), the graph goes up (\(y\to+\infty\)) and as \(x\to+\infty\), the graph goes down (\(y\to-\infty\)), so the degree is odd and the leading coefficient is negative. So degree 3? But the graph has a touch point (multiplicity 2) and a cross point (multiplicity 1), so total degree 3. So the function is \(y=a(x + 6)^2(x - r)\), where \(r\) is the cross root. Wait, the cross root: looking at the graph, it crosses the x - axis at \(x = 0\)? No, the y - intercept is at \((0,-30)\), so it crosses the y - axis at \((0,-30)\) and crosses the x - axis at \(x=-6\) (touch) and \(x = 0\) (cross)? Wait, no, the x - axis crossing: the graph touches at \(x=-6\) and crosses at \(x = 0\) and maybe \(x=-1\). Wait, let's start over. Let's assume the function is a cubic function with a double root at \(x=-6\) and a single root at \(x = 0\). So \(y=a(x + 6)^2x\). Now, find \(a\). When \(x = 0\), \(y = 0\), but the y - intercept is \((0,-30)\), so that's not correct. Wait, maybe the function is \(y=a(x + 6)^2(x)(x - 1)\)? No, let's look at the y - intercept. The y - intercept is \((0,-30)\). Let's assume the function is \(y=-\frac{5}{36}(x + 6)^2x(x + 1)\). Wait, no, let's try a simpler approach. The graph touches the x - axis at \(x=-6\) (so \((x + 6)^2\)) and crosses at \(x = 0\) and \(x=-1\). Let's take the function \(y=a(x + 6)^2x(x + 1)\). Plug in \(x = 0\), \(y=-30\): \(-30=a(36)(0)(1)\), which is 0. So that's wrong. Maybe the root at \(x=-6\) (multiplicity 2), \(x = 0\) (multiplicity 1), and the function is \(y=a(x + 6)^2x\). But when \(x = 0\), \(y = 0\), but the y - intercept is \(-30\). So maybe the graph has a root at \(x=-6\) (multiplicity 2), \(x = 0\) (multiplicity 1), and the leading coefficient is such that when \(x = 0\), \(y=-30\). Wait, no, if \(y=a(x + 6)^2x\), then at \(x = 0\), \(y = 0\). So my initial assumption about the roots is wrong. Let's look at the graph again. The graph crosses the x - axis at \(x = 0\) and touches at \(x=-6\). Let's assume the function is \(y=a(x + 6)^2x\). Now, we need to find \(a\) such that when \(x = 0\), \(y=-30\), but that's not possible. Wait, maybe the graph is a quartic function. Let's assume the function is \(y=a(x + 6)^2x(x + 1)\). The degree is 4 (even), but the end - behavior: as \(x\to-\infty\), \(y\to+\infty\) and \(x\to+\infty\), \(y\to-\infty\) is for odd degree. So that's a contradiction. Wait, I think I messed up the end - behavior. As \(x\to-\infty\), the graph goes up (positive) and as \(x\to+\infty\), it goes down (negative), so the leading coefficient is negative and the degree is odd. So degree 3. So the function has a double root (multiplicity 2) and a single root (multiplicity 1), so total degree 3. Let the double root be at \(x=-6\) and the single root be at \(x = 0\). So \(y=a(x + 6)^2x\). Now, we need to find a point on the graph other than the intercepts. Let's pick a point. When \(x=-6\), \(y = 0\) (touching). When \(x=-5\), let's see the graph. At \(x=-5\), the graph is above the x - axis. Plug \(x=-5\) into \(y=a(x + 6)^2x\): \(y=a(-5 + 6)^2(-5)=a(1)^2(-5)=-5a\). Since it's above the x - axis, \(y>0\), so \(-5a>0\) implies \(a<0\). Now, the y - intercept is \((0,-30)\), but \(y=a(x + 6)^2x\) at \(x = 0\) is 0. So this is wrong. Wait, maybe the single root is at \(x = 1\)? No, the graph crosses the y - axis at \((0,-30)\) and goes down. Wait, let's try a different approach. Let's assume the function is \(y=-5(x + 6)^2x\). At \(x = 0\), \(y = 0\), not - 30. \(y=-\frac{5}{36}(x + 6)^2x(x + 1)\): at \(x = 0\), \(y = 0\). Wait, maybe the graph has a root at \(x=-6\) (multiplicity 2), \(x = 0\) (multiplicity 1), and the function is \(y=a(x + 6)^2x\), and we made a mistake in the y - intercept. Wait, the y - intercept is \((0,-30)\), so when \(x = 0\), \(y=-30\). So \( - 30=a(0 + 6)^2(0)\), which is 0. So this is impossible. Therefore, the graph must have a root at \(x=-6\) (multiplicity 2), \(x = 0\) (multiplicity 1), and \(x=-1\) (multiplicity 1), so the function is \(y=a(x + 6)^2x(x + 1)\). Now, when \(x = 0\), \(y = 0\), but the y - intercept is \((0,-30)\), so there must be a mistake. Wait, maybe the graph is a cubic function with a double root at \(x=-6\) and a single root at \(x=-1\), and the y - intercept is \((0,-30)\). So \(y=a(x + 6)^2(x + 1)\). Plug in \(x = 0\): \(y=a(6)^2(1)=36a\). We know \(y=-30\) when \(x = 0\), so \(36a=-30\), \(a=-\frac{30}{36}=-\frac{5}{6}\). So the function is \(y=-\frac{5}{6}(x + 6)^2(x + 1)\). Let's check the end - behavior: degree 3, leading coefficient negative, so as \(x\to-\infty\), \(y\to+\infty\) and \(x\to+\infty\), \(y\to-\infty\), which matches. Let's check the root at \(x=-6\) (double root, touches x - axis) and \(x=-1\) (single root, crosses x - axis). This seems to fit.

Step2: Verify the function

Let's expand \(y=-\frac{5}{6}(x + 6)^2(x + 1)\). First, \((x + 6)^2=x^{2}+12x + 36\). Then, multiply by \((x + 1)\): \((x^{2}+12x + 36)(x + 1)=x^{3}+13x^{2}+48x + 36\). Then multiply by \(-\frac{5}{6}\): \(y=-\frac{5}{6}x^{3}-\frac{65}{6}x^{2}-40x - 30\). When \(x = 0\), \(y=-30\), which matches the y - intercept. The root at \(x=-6\) (double root) and \(x=-1\) (single root) matches the x - intercepts (touch at \(x=-6\), cross at \(x=-1\)). The end - behavior also matches.

Answer:

\(y = -\frac{5}{6}(x + 6)^2(x + 1)\) (or equivalent factored or expanded form)