Question

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

Explanation:

Step1: Identify roots of the function

The graph intersects the x - axis at \(x=-8\), and touches the x - axis at \(x = 4\) and \(x=6\) (since it has a double root at the point where it touches the x - axis, so the multiplicity of the root at \(x = 4\) and \(x=6\) is 2? Wait, no, when the graph touches the x - axis at a single point between 4 and 6? Wait, looking at the graph, the x - intercepts: one at \(x=-8\), and a double root at \(x = 5\)? Wait, no, the graph touches the x - axis between \(x = 4\) and \(x=6\), let's assume the roots are \(x=-8\), \(x = 5\) (with multiplicity 2) or more accurately, from the graph, the roots are \(x=-8\), and a repeated root at \(x = 5\) (since the graph touches the x - axis at \(x = 5\)). Wait, maybe the roots are \(x=-8\), \(x = 4\) and \(x=6\)? No, the graph touches the x - axis at a single point between 4 and 6, so it's a double root. Let's assume the function is a cubic? No, the graph has a "w" shape? Wait, no, the leading coefficient: as \(x\to\infty\), \(y\to\infty\) and as \(x\to-\infty\), \(y\to-\infty\), so the leading coefficient is positive and the degree is 3? Wait, no, the graph has two turning points (a local maximum and a local minimum), so the degree is at least 3. Wait, let's re - examine the x - intercepts. The graph crosses the x - axis at \(x=-8\) and touches the x - axis at \(x = 5\) (a double root). So the function can be written in factored form as \(y=a(x + 8)(x - 5)^2\).

Step2: Find the value of \(a\)

We know that the graph passes through the y - axis at \((0,150)\) (wait, looking at the graph, when \(x = 0\), \(y\) is around 150? Wait, the y - intercept: from the graph, when \(x = 0\), \(y\) is 150? Wait, let's check the graph again. The y - axis is at \(x = 0\), and the graph passes through \((0,150)\)? Wait, no, the grid: each square is, let's assume the y - axis at \(x = 0\), the point on the graph is at \(y = 150\)? Wait, let's use the factored form \(y=a(x + 8)(x - 5)^2\). Plug in \(x = 0\), \(y=a(0 + 8)(0 - 5)^2=a\times8\times25 = 200a\). From the graph, when \(x = 0\), \(y=150\)? Wait, no, maybe my initial assumption of the root is wrong. Wait, the graph intersects the x - axis at \(x=-8\), and has a double root at \(x = 5\), and when \(x = 0\), \(y = 150\)? Wait, no, let's recalculate. Wait, maybe the roots are \(x=-8\), \(x = 4\) and \(x=6\), but the graph touches the x - axis at \(x = 5\) (mid - point of 4 and 6). Let's try \(y=a(x + 8)(x - 4)(x - 6)\). Now, find \(a\) using the y - intercept. When \(x = 0\), \(y=a(0 + 8)(0 - 4)(0 - 6)=a\times8\times(-4)\times(-6)=a\times192\). From the graph, when \(x = 0\), \(y = 150\)? Wait, no, the graph at \(x = 0\) is at \(y = 150\)? Wait, the y - axis: the grid lines, 500, 400, 300, 200, 100, 0, - 100, etc. The point on the graph at \(x = 0\) is at \(y = 150\)? Wait, maybe \(a=\frac{150}{192}=\frac{25}{32}\)? No, that seems messy. Wait, maybe the roots are \(x=-8\), \(x = 5\) (double root). Let's try \(y=a(x + 8)(x - 5)^2\). When \(x = 0\), \(y=a\times8\times25=200a\). If we assume that at \(x = 0\), \(y = 150\), then \(200a=150\), so \(a=\frac{3}{4}\). Wait, but let's check the shape. Alternatively, maybe the function is \(y=\frac{1}{2}(x + 8)(x - 5)^2\). Wait, no, let's start over.

Wait, the graph has x - intercepts at \(x=-8\) and a double root at \(x = 5\) (since it touches the x - axis at \(x = 5\)). So the factored form is \(y=a(x + 8)(x - 5)^2\). Now, let's find \(a\). Let's take a point on the graph. When \(x = 0\), \(y\) is 150 (from the graph, the y - intercept is 150). So:

\(150=a(0 + 8)(0 - 5)^2\)

\(150=a\times8\times25\)

\(150 = 200a\)

\(a=\frac{150}{200}=\frac{3}{4}\)

Wait, but let's check the local maximum and minimum. Alternatively, maybe the roots are \(x=-8\), \(x = 4\) and \(x=6\), so the function is \(y=a(x + 8)(x - 4)(x - 6)\). When \(x = 0\), \(y=a\times8\times(-4)\times(-6)=192a\). If \(y = 150\) when \(x = 0\), then \(a=\frac{150}{192}=\frac{25}{32}\approx0.78125\). But maybe the y - intercept is 160? Let's assume \(a=\frac{1}{12}\), no. Wait, let's look at the graph again. The y - intercept: the graph passes through \((0,150)\)? Wait, the grid: each square on the y - axis is 100? Wait, the y - axis has marks at 100, 200, 300, etc. So when \(x = 0\), the graph is at \(y = 150\)? Wait, no, the point on the graph at \(x = 0\) is between 100 and 200, closer to 150. Let's use the factored form with roots \(x=-8\), \(x = 5\) (double root). So \(y=a(x + 8)(x - 5)^2\). Let's expand this: \(y=a(x + 8)(x^{2}-10x + 25)=a(x^{3}-10x^{2}+25x+8x^{2}-80x + 200)=a(x^{3}-2x^{2}-55x + 200)\). Now, if we take \(a = \frac{3}{4}\), then \(y=\frac{3}{4}(x^{3}-2x^{2}-55x + 200)=\frac{3}{4}x^{3}-\frac{3}{2}x^{2}-\frac{165}{4}x + 150\). Let's check the x - intercept at \(x=-8\): \(\frac{3}{4}(-8)^{3}-\frac{3}{2}(-8)^{2}-\frac{165}{4}(-8)+150=\frac{3}{4}(-512)-\frac{3}{2}(64)+\frac{1320}{4}+150=-384 - 96+330 + 150=0\). Good. At \(x = 5\): \(\frac{3}{4}(125)-\frac{3}{2}(25)-\frac{165}{4}(5)+150=\frac{375}{4}-\frac{75}{2}-\frac{825}{4}+150=\frac{375 - 150 - 825+600}{4}=\frac{0}{4}=0\). Good. And the multiplicity at \(x = 5\) is 2, so the graph touches the x - axis there.

Alternatively, a simpler approach: the roots are \(x=-8\), \(x = 5\) (double root), so the function can be written as \(y=\frac{3}{4}(x + 8)(x - 5)^2\) or we can adjust \(a\) to make the y - intercept correct. Wait, when \(x = 0\), \(y=\frac{3}{4}(8)(25)=150\), which matches the y - intercept (assuming the y - intercept is 150).

Step3: Write the function

So the function in factored form is \(y=\frac{3}{4}(x + 8)(x - 5)^2\). We can also expand it, but the factored form is acceptable.

Answer:

\(y=\frac{3}{4}(x + 8)(x - 5)^2\) (or an equivalent expanded form, or with a different value of \(a\) if the y - intercept is estimated differently. Another possible form: if we consider the roots as \(x=-8\), \(x = 4\) and \(x=6\), then \(y=\frac{5}{64}(x + 8)(x - 4)(x - 6)\) (when \(x = 0\), \(y=\frac{5}{64}(8)(-4)(-6)=\frac{5}{64}\times192 = 15\), no, that's not correct. So the first form with \(a=\frac{3}{4}\) is better.)