Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Ask your questions and receive precise answers from experienced professionals across different disciplines. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Given that f(x) is a cubic function with zeros at -3, 0, and 7, find an equation for f(x) given that f(-8) = -5.

Sagot :

gmany

Answer:

[tex]\huge\boxed{f(x)=\dfrac{1}{120}x^3-\dfrac{1}{30}x^2-\dfrac{7}{40}x}[/tex]

Step-by-step explanation:

[tex]f(x)=a(x+3)(x-0)(x-7)=ax(x+3)(x-7)\\\\=ax\bigg((x)(x)+(x)(-7)+(3)(x)+(3)(-7)\bigg)\\\\=ax(x^2-7x+3x-21)=ax(x^2-4x-21)\\\\=(ax)(x^2)+(ax)(-4x)+(ax)(-21)\\\\=ax^3-4ax^2-21ax\qquad(*)\\\\f(-8)=-5[/tex]

[tex]\text{substitute}\ x=-8\\\\f(-8)=(a)(-8)^3-(4a)(-8)^2-(21)(-8)a=-512a-(4a)(64)+168a\\\\=-512a-256a+168a=-600a\\\\f(-8)=-5\\\\\text{therefore}\\\\-600a=-5\qquad|\text{divide both sides by (-600)}\\\\a=\dfrac{-5}{-600}\\\\a=\dfrac{1}{120}[/tex]

[tex]\text{Substitute to}\ (*):\\\\f(x)=\dfrac{1}{120}x^3-4\cdot\dfrac{1}{120}x^2-21\cdot\dfrac{1}{120}x=\dfrac{1}{120}x^3-\dfrac{1}{30}x^2-\dfrac{7}{40}x[/tex]

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.