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

 

Please explain very well in this question. I am so sorry! I just need a little help.

Guest Dec 4, 2014

Best Answer 

 #1
avatar+18827 
+10

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

$$\small{\text{The polynomial f(x) of degree 3 is }} f(x) = ax^3+bx^2+cx+d$$

I.  We need a, b, c and d :

$$\small{\text{
\begin{array}{r|r|lrclrclccl}
\hline
x & y & &f(x)& =& ax^3+bx^2+cx+d & && &\textcolor[rgb]{1,0,0}{d}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{0} \\
\hline
-1 & 15& (1) & 15 &=& a(-1)^3+b(-1)^2+c(-1) +d & 15&=& -a+b-c+d & 15&=&-a+b-c\\
0 & 0 & (2) & 0 &=& a(0)^3+b(0)^2+c(0) +d & \textcolor[rgb]{1,0,0}{0} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{d} & -5&=&a+b+c\\
1 & -5 & (3) & -5 &=& a(1)^3+b(1)^2+c(1) +d & -5 &=& a+b+c+d & \\
2 & 12 & (4) & 12 &=& a(2)^3+b(2)^2+c(2) +d & 12 &=& 8a+4b+2c+d & 12 &=& 8a+4b+2c\\
\hline
\end{array}
}}$$

d=0:

(1) -a + b - c = 15

(2)  a + b + c = -5

(4) 8a+4b+2c = 12 | :2    $$\Rightarrow$$   (4) 4a + 2b + c = 6

----------------------------------------------------------

(1)+(2): 2b = 10  $$\Rightarrow$$  $$\textcolor[rgb]{1,0,0}{ b = 5}$$

----------------------------------------------------------

b=5:

(1)   a + c = -10

(2)   a + c = -10

(4) 4a + c =  -4

----------------------------------------------------------

(4)-(2): 3a = -4 -(-10) = 6  $$\Rightarrow$$  3a = -4+10  $$\Rightarrow$$  3a = 6   =>  $$\textcolor[rgb]{1,0,0}{a=2}$$ 

(1)  2 + c = -10  $$\Rightarrow$$  $$\textcolor[rgb]{1,0,0}{c = -12}$$

$$\small{\text{The polynomial f(x) of degree 3 is }} f(x) = 2x^3+5x^2-12x+0$$

 

II.  x-intercepts of the graph of $$f(x)$$?

 

$$2x^3+5x^2-12x = 0 \\
\underbrace{x}_{=0}*\underbrace{( 2x^2+5x-12 )}_{=0} = 0 \\\\
\textcolor[rgb]{1,0,0}{x_1 = 0} \\\\
2x^2+5x-12 = 0 \quad | \quad\textcolor[rgb]{0,0,1}{ ax^2+bx+c=0 => x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} } \\\\
x_{2,3}=\frac{-5\pm\sqrt{25-4*2*(-12)} }{4} \\\\
x_{2,3}=\frac{-5\pm\sqrt{121} }{4} \\\\
x_{2,3}=\frac{-5\pm\11 }{4} \\\\
x_2=\frac{-5+11 }{4} = \frac{6}{4} = 1.5 \quad \Rightarrow \quad \textcolor[rgb]{1,0,0}{x_2=1.5} \\\\
x_3=\frac{-5-11 }{4} = \frac{-16}{4} = -4 \quad \Rightarrow \quad \textcolor[rgb]{1,0,0}{x_3=-4} \\\\$$

The x-intercepts are: -4, 0 and 1.5

heureka  Dec 4, 2014
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1+0 Answers

 #1
avatar+18827 
+10
Best Answer

----------------------------------------------------------

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

$$\small{\text{The polynomial f(x) of degree 3 is }} f(x) = ax^3+bx^2+cx+d$$

I.  We need a, b, c and d :

$$\small{\text{
\begin{array}{r|r|lrclrclccl}
\hline
x & y & &f(x)& =& ax^3+bx^2+cx+d & && &\textcolor[rgb]{1,0,0}{d}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{0} \\
\hline
-1 & 15& (1) & 15 &=& a(-1)^3+b(-1)^2+c(-1) +d & 15&=& -a+b-c+d & 15&=&-a+b-c\\
0 & 0 & (2) & 0 &=& a(0)^3+b(0)^2+c(0) +d & \textcolor[rgb]{1,0,0}{0} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{d} & -5&=&a+b+c\\
1 & -5 & (3) & -5 &=& a(1)^3+b(1)^2+c(1) +d & -5 &=& a+b+c+d & \\
2 & 12 & (4) & 12 &=& a(2)^3+b(2)^2+c(2) +d & 12 &=& 8a+4b+2c+d & 12 &=& 8a+4b+2c\\
\hline
\end{array}
}}$$

d=0:

(1) -a + b - c = 15

(2)  a + b + c = -5

(4) 8a+4b+2c = 12 | :2    $$\Rightarrow$$   (4) 4a + 2b + c = 6

----------------------------------------------------------

(1)+(2): 2b = 10  $$\Rightarrow$$  $$\textcolor[rgb]{1,0,0}{ b = 5}$$

----------------------------------------------------------

b=5:

(1)   a + c = -10

(2)   a + c = -10

(4) 4a + c =  -4

----------------------------------------------------------

(4)-(2): 3a = -4 -(-10) = 6  $$\Rightarrow$$  3a = -4+10  $$\Rightarrow$$  3a = 6   =>  $$\textcolor[rgb]{1,0,0}{a=2}$$ 

(1)  2 + c = -10  $$\Rightarrow$$  $$\textcolor[rgb]{1,0,0}{c = -12}$$

$$\small{\text{The polynomial f(x) of degree 3 is }} f(x) = 2x^3+5x^2-12x+0$$

 

II.  x-intercepts of the graph of $$f(x)$$?

 

$$2x^3+5x^2-12x = 0 \\
\underbrace{x}_{=0}*\underbrace{( 2x^2+5x-12 )}_{=0} = 0 \\\\
\textcolor[rgb]{1,0,0}{x_1 = 0} \\\\
2x^2+5x-12 = 0 \quad | \quad\textcolor[rgb]{0,0,1}{ ax^2+bx+c=0 => x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} } \\\\
x_{2,3}=\frac{-5\pm\sqrt{25-4*2*(-12)} }{4} \\\\
x_{2,3}=\frac{-5\pm\sqrt{121} }{4} \\\\
x_{2,3}=\frac{-5\pm\11 }{4} \\\\
x_2=\frac{-5+11 }{4} = \frac{6}{4} = 1.5 \quad \Rightarrow \quad \textcolor[rgb]{1,0,0}{x_2=1.5} \\\\
x_3=\frac{-5-11 }{4} = \frac{-16}{4} = -4 \quad \Rightarrow \quad \textcolor[rgb]{1,0,0}{x_3=-4} \\\\$$

The x-intercepts are: -4, 0 and 1.5

heureka  Dec 4, 2014

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