Question

e a function in any form that would match the graph shown below.
answer attempt 1 out of 2
f(x) =

Explanation:

Step1: Identify roots and multiplicity

The graph touches the x - axis at \(x = 2\) (so multiplicity 2) and crosses at \(x=6\). The y - intercept is at \((0,18)\) (approx, but let's assume we can use factored form. The general form for a polynomial with roots \(r_1\), \(r_2\),... with multiplicities \(m_1\), \(m_2\),... is \(f(x)=a(x - r_1)^{m_1}(x - r_2)^{m_2}\cdots\). So here, \(f(x)=a(x - 2)^{2}(x - 6)\).

Step2: Find the leading coefficient \(a\)

Use the y - intercept \((0,y)\). When \(x = 0\), \(y=a(0 - 2)^{2}(0 - 6)=a\times4\times(- 6)=-24a\). From the graph, when \(x = 0\), \(y = 18\) (close to the given graph's y - intercept around 18). So \(-24a=18\), \(a=-\frac{18}{24}=-\frac{3}{4}\). Wait, but maybe the y - intercept is 18? Wait, the graph at \(x = 0\) is at \(y = 18\) (looking at the grid, the y - axis has 16, 12, etc. So let's recalculate. Wait, if we take the y - intercept as \(y = 18\), then \(-24a=18\), \(a=-\frac{3}{4}\). But maybe a better approach: let's check the graph again. Alternatively, maybe the y - intercept is 18, so \(f(0)=a( - 2)^{2}( - 6)=-24a = 18\), so \(a=-\frac{3}{4}\). Then \(f(x)=-\frac{3}{4}(x - 2)^{2}(x - 6)\). Let's expand it: \((x - 2)^{2}=x^{2}-4x + 4\), then \((x^{2}-4x + 4)(x - 6)=x^{3}-6x^{2}-4x^{2}+24x + 4x-24=x^{3}-10x^{2}+28x - 24\). Then \(f(x)=-\frac{3}{4}x^{3}+\frac{30}{4}x^{2}-\frac{84}{4}x + 18=-\frac{3}{4}x^{3}+\frac{15}{2}x^{2}-21x + 18\). Alternatively, maybe the y - intercept is 18, so another way: let's assume the function is a cubic (since it has two turning points, degree 3). The roots are \(x = 2\) (double root) and \(x = 6\). So factored form \(f(x)=a(x - 2)^{2}(x - 6)\). At \(x = 0\), \(f(0)=a(4)(-6)=-24a\). From the graph, when \(x = 0\), \(y = 18\), so \(a=-\frac{18}{24}=-\frac{3}{4}\). So \(f(x)=-\frac{3}{4}(x - 2)^{2}(x - 6)\). Let's check \(x = 2\): \(f(2)=0\), correct (touches x - axis). \(x = 6\): \(f(6)=0\), correct (crosses x - axis). The leading coefficient is negative, so as \(x
ightarrow\infty\), \(f(x)
ightarrow-\infty\) and as \(x
ightarrow-\infty\), \(f(x)
ightarrow\infty\), which matches the graph (right end goes down, left end goes up? Wait, no: for a cubic with leading coefficient negative, as \(x
ightarrow\infty\), \(f(x)
ightarrow-\infty\), as \(x
ightarrow-\infty\), \(f(x)
ightarrow\infty\). The graph on the right (x large positive) goes down, left (x large negative) goes up? Wait, the graph is on the right side (x>0) mostly. Wait, maybe my root at \(x = 6\) is correct. Alternatively, maybe the root is \(x = 6\) and \(x = 2\) (double root). So the function is \(f(x)=a(x - 2)^{2}(x - 6)\) with \(a=-\frac{3}{4}\) (from y - intercept).

Answer:

\(f(x)=-\frac{3}{4}(x - 2)^{2}(x - 6)\) (or expanded form)