The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12 , then what are the $x$-intercepts of the graph of f? Explain.

Guest Aug 13, 2018

edited by
Guest
Aug 13, 2018

edited by Guest Aug 14, 2018

edited by Guest Aug 14, 2018

#1**+2 **

I am here to help you learn, not to help you meet your deadlines.

The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12 , then what are the $x$-intercepts of the graph of f? Explain.

\(f(x)=ax^3+bx^2+cx+d\\ f(0)=d=0 \qquad \text{so d =0}\\~\\ f(-1)=-a+b-c=15\qquad (1a)\\ f(1)=a+b+c=-5\qquad (2a)\\ f(2)=8a+4b+2c=12\qquad(3a)\\~\\ b-15=a+c\qquad (1b)\\ a+c=-b-5\qquad (2a)\\ so\\ b-15=-b-5\\ 2b=10\\ b=5\\ so\ \)

\(-a+5-c=15\qquad (1c)\\ a+c=-10 \qquad(1d)\\ 8a+20+2c=12\qquad(3c)\\ 4a+c=2\qquad(3d)\\\)

\(a+c=-10 \qquad(1d)\\ 4a+c=2\qquad(3d)\\ Solve\;\; simultaneously\\ etc\)

Please let this guest finish it and check for errors all by him/herself !

Melody Aug 14, 2018

#2**+1 **

So for this do i just solve for a and c next? and then solve the quadratic to find the answer?

Guest Aug 14, 2018

#3**+1 **

Mmm lets see.

\(a+c=-10\\ 4a+c=2\\ so\\ 3a=12\\ a=4\\ c=-10-4=-14\\\)

So we have

\(f(x)=ax^3+bx^2+cx+d\\ f(x)=ax^3+5x^2+cx\\ f(x)=4x^3+5x^2-14x\\ f(x)=x(4x^2+5x-14)\\ F(x)=0\;\; when \;\; x(4x^2+5x-14)=0\\ x=0\;\;or\;\;4x^2+5x-14=0\\ x=0\;\;or\;\; x=\frac{-5\pm \sqrt{25+224}}{8}\\ x=0\;\;or\;\; x=\frac{-5\pm \sqrt{249}}{8}\\ \)

If there are no careless errors in my working then these are the x intercepts.

You should check my answer by substituting the answers in for x and seeing if the answer for f(x) is zero.

Melody
Aug 15, 2018