ANSWER
f(x) = -18x³ + 21x² + 10x - 8
EXPLANATION
As we can see, the x-intercepts are x = -2/3, x = 1/2 and x = 4/3. Thus, these are the zeros of the polynomial.
Given a polynomial with zeros x₁, x₂, and x₃, the polynomial's degree is 3 and it can be written with the factors,
[tex]f(x)=a(x-x_1)(x-x_2)(x-x_3)[/tex]Where a is a coefficient given by the location of the y-intercept.
With the zeros, we have the function,
[tex]f(x)=a\mleft(x+\frac{2}{3}\mright)\mleft(x-\frac{1}{2}\mright)\mleft(x-\frac{4}{3}\mright)[/tex]In the graph, the y-intercept is shown: (0, -8). This means that when x is 0, f(0) is -8. This is what we have to use to find the coefficient a,
[tex]-8=a\mleft(0+\frac{2}{3}\mright)\mleft(0-\frac{1}{2}\mright)\mleft(0-\frac{4}{3}\mright)[/tex]We have the product,
[tex]-8=a\mleft(\frac{2}{3}\mright)\mleft(-\frac{1}{2}\mright)\mleft(-\frac{4}{3}\mright)[/tex]Solve the product,
[tex]\begin{gathered} -8=a\mleft(\frac{2(-1)(-4)}{3\cdot2\cdot3}\mright) \\ -8=a\mleft(\frac{4}{9}\mright) \end{gathered}[/tex]To find a, multiply both sides by 9,
[tex]\begin{gathered} -8\cdot9=a\cdot\frac{4}{9}\cdot9 \\ -72=4a \end{gathered}[/tex]And divide both sides by 4,
[tex]\begin{gathered} -\frac{72}{4}=\frac{4a}{4} \\ -18=a \end{gathered}[/tex]Thus the coefficient a is -18, and the function is,
[tex]f(x)=-18\mleft(x+\frac{2}{3}\mright)\mleft(x-\frac{1}{2}\mright)\mleft(x-\frac{4}{3}\mright)[/tex]Now we have to write it in standard form. To do so, we have to multiply the factors to get a function in the form,
[tex]f(x)=ax^3+bx^2+cx+d[/tex]Apply the distributive property to the last two factors,
[tex]\begin{gathered} f(x)=-18\mleft(x+\frac{2}{3}\mright)\mleft(x\cdot x-x\cdot\frac{4}{3}-x\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{4}{3}\mright) \\ \\ f(x)=-18\mleft(x+\frac{2}{3}\mright)\mleft(x^2-\frac{11}{6}x+\frac{2}{3}\mright) \end{gathered}[/tex]Then do the same with the first factor,
[tex]\begin{gathered} f(x)=-18\mleft(x^2\cdot x-\frac{11}{6}x\cdot x+\frac{2}{3}x+\frac{2}{3}x^2-\frac{11}{6}\cdot\frac{2}{3}x+\frac{2}{3}\cdot\frac{2}{3}\mright) \\ \\ f(x)=-18\mleft(x^3-\frac{11}{6}x^2+\frac{2}{3}x^2+\frac{2}{3}x-\frac{11}{9}x+\frac{4}{9}\mright) \end{gathered}[/tex]Add like terms,
[tex]f(x)=-18\mleft(x^3-\frac{7}{6}x^2-\frac{5}{9}x+\frac{4}{9}\mright)[/tex]And finally, multiply each term by -18,
[tex]\begin{gathered} f(x)=-18x^3+18\cdot\frac{7}{6}x^2+18\cdot\frac{5}{9}x-18\cdot\frac{4}{9} \\ f(x)=-18x^3+21x^2+10x-8 \end{gathered}[/tex]Hence, the polynomial function whose graph is in the question is f(x) = -18x³ + 21x² + 10x - 8
