Question

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

Explanation:

Step1: Identify roots and multiplicities

The graph touches the x - axis at \(x = - 8\) and \(x = 5\) (wait, looking at the graph, the x - intercepts where the graph touches the x - axis: at \(x=-8\) (a double root, since the graph touches and turns) and \(x = 5\)? Wait, no, looking at the graph again, the x - intercepts: one at \(x=-8\) (touching, so even multiplicity), one at \(x = 5\) (touching, even multiplicity)? Wait, no, the graph crosses or touches? Wait, the left part: at \(x=-8\), the graph touches the x - axis (so multiplicity 2), then at \(x = 5\), the graph touches the x - axis (multiplicity 2)? Wait, no, maybe I misread. Wait, the graph: let's check the x - intercepts. The graph touches the x - axis at \(x=-8\) (so a root with even multiplicity, say 2) and at \(x = 5\) (wait, no, the right side: at \(x = 5\)? Wait, the grid: the x - axis has marks at - 10, - 8, - 6, - 4, - 2, 0, 2, 4, 6, 8, 10. The graph touches the x - axis at \(x=-8\) (so a root \(x=-8\), multiplicity 2) and at \(x = 5\)? Wait, no, the right - hand touch is at \(x = 5\)? Wait, no, looking at the graph, the right - hand touch is at \(x = 5\)? Wait, maybe \(x = 5\) is a root with multiplicity 2, and also, is there another root? Wait, no, maybe the function is a quartic? Wait, no, let's think about the general form. A function that touches the x - axis at \(x = a\) and \(x = b\) has factors \((x - a)^2\) and \((x - b)^2\). Wait, but the graph also has a y - intercept at \((0,-800)\). Let's assume the roots are \(x=-8\) (multiplicity 2) and \(x = 5\) (multiplicity 2)? Wait, no, maybe \(x=-8\) (multiplicity 2) and \(x = 5\) (multiplicity 2), and the leading coefficient? Wait, let's try to write the function as \(y=a(x + 8)^2(x - 5)^2\). Now, use the y - intercept \((0,-800)\). Plug in \(x = 0\), \(y=-800\):

\(-800=a(0 + 8)^2(0 - 5)^2\)

\(-800=a\times64\times25\)

\(-800=a\times1600\)

\(a=\frac{-800}{1600}=-\frac{1}{2}\)

So the function is \(y =-\frac{1}{2}(x + 8)^2(x - 5)^2\). Let's expand this to check.

First, \((x + 8)^2=x^{2}+16x + 64\), \((x - 5)^2=x^{2}-10x + 25\)

Multiply them: \((x^{2}+16x + 64)(x^{2}-10x + 25)=x^{4}-10x^{3}+25x^{2}+16x^{3}-160x^{2}+400x + 64x^{2}-640x + 1600\)

Simplify: \(x^{4}+6x^{3}-71x^{2}-240x + 1600\)

Then multiply by \(-\frac{1}{2}\): \(y=-\frac{1}{2}x^{4}-3x^{3}+\frac{71}{2}x^{2}+120x - 800\)

Wait, but let's check the y - intercept: when \(x = 0\), \(y=-800\), which matches. Let's check the shape. The leading coefficient is negative (\(-\frac{1}{2}\)), so as \(x\to\pm\infty\), \(y\to-\infty\), which matches the graph (left end goes down, right end goes down). The graph touches the x - axis at \(x=-8\) and \(x = 5\) (since the factors are squared), which matches the graph touching the x - axis at those points.

Step2: Verify the function

We assumed the roots are \(x=-8\) (multiplicity 2) and \(x = 5\) (multiplicity 2), found the leading coefficient \(a =-\frac{1}{2}\), and the function \(y =-\frac{1}{2}(x + 8)^2(x - 5)^2\) satisfies the y - intercept and the end - behavior and the x - intercepts (touching, so even multiplicity).

Answer:

\(y =-\frac{1}{2}(x + 8)^2(x - 5)^2\) (or expanded form \(y=-\frac{1}{2}x^{4}-3x^{3}+\frac{71}{2}x^{2}+120x - 800\))