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 $f$?
+26367 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 $f$?
The polynomial of degree 3 is:
\(f(x) = ax^3+bx^2+cx+d\)
\(f(0) = 0:\)
\(\begin{array}{|rcll|} \hline f(0) = 0 &=& a\cdot 0^3+b\cdot 0^2+c\cdot 0+d \\ 0 &=& d \\ \mathbf{d} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)
The polyomial of degree 3 is now:
\(f(x) = ax^3+bx^2+cx\)
\(f(-1) = 15:\)
\(\begin{array}{|rcll|} \hline f(-1) = 15 &=& a\cdot (-1)^3+b\cdot (-1)^2+c\cdot (-1) \\ 15 &=& -a +b-c \\ \mathbf{ -a+b-c} &\mathbf{=}& \mathbf{15} \qquad (1) \\ \hline \end{array}\)
\(f(1) = -5:\)
\(\begin{array}{|rcll|} \hline f(1) = -5 &=& a\cdot 1^3+b\cdot 1^2+c\cdot 1 \\ -5 &=& a +b+c \\ \mathbf{ a +b+c } &\mathbf{=}& \mathbf{-5} \qquad (2) \\ \hline \end{array}\)
\(f(2) = 12:\)
\(\begin{array}{|rcll|} \hline f(2) = 12 &=& a\cdot 2^3+b\cdot 2^2+c\cdot 2 \\ 12&=& 8a+4b+2c \quad & | \quad : 2 \\ \mathbf{ 4a +2b+c } &\mathbf{=}& \mathbf{6} \qquad (3) \\ \hline \end{array}\)
find a,b,c:
\(\begin{array}{|lrcll|} \hline (1) & \mathbf{ -a+b-c} &\mathbf{=}& \mathbf{15} \\ (2) & \mathbf{ a +b+c } &\mathbf{=}& \mathbf{-5} \\ (3) & \mathbf{ 4a +2b+c } &\mathbf{=}& \mathbf{6} \\ \hline (1)+(2): & 2b &=& 10 \\ & \mathbf{b} &\mathbf{=}& \mathbf{5} \\ \hline (3)-(2): & 3a+b &=& 11 \quad & | \quad b=5 \\ & 3a+5 &=& 11 \\ & 3a &=& 6 \\ & \mathbf{a} &\mathbf{=}& \mathbf{2} \\ \hline (2) & a +b+c & = & -5 \quad & | \quad a=2\quad b=5 \\ & 2 +5+c & = & -5 \\ & 7+c & = & -5 \\ & c & = & -5 -7 \\ & \mathbf{c} &\mathbf{=}& \mathbf{-12} \\ \hline \end{array}\)
The polyomial of degree 3 is:
\(f(x) = 2x^3+5x^2-12x\)
The graph:
\(\text{what are the $x$-intercepts of the graph of $f\ $ ?}\)
\(\begin{array}{|lrcll|} \hline & 2x^3+5x^2-12x &=& 0 \\ & x(2x^2+5x-12) &=& 0 \\ \hline 1. & \mathbf{x_1} &\mathbf{=}& \mathbf{0} \\ \hline 2. \\ & 2x^2+5x-12 &=& 0 \\ & x &=& \dfrac{-5 \pm \sqrt{25-4\cdot 2 \cdot (-12) } } {2\cdot 2} \\ & &=& \dfrac{-5 \pm \sqrt{121} } {4} \\ & &=& \dfrac{-5 \pm 11 } {4} \\\\ & x_2 &=& \dfrac{-5 + 11 } {4} \\ & \mathbf{x_2} &\mathbf{=}& \mathbf{1.5} \\\\ & x_3 &=& \dfrac{-5 - 11 } {4} \\ & \mathbf{x_3} &\mathbf{=}& \mathbf{-4} \\ \hline \end{array}\)
\(\text{ The $x$-intercepts of the graph of $f$ are x=-4, x=0, and x = 1.5 }\)
