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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?

 

 

Thank you all so much!

AnonymousConfusedGuy  Nov 28, 2017

Best Answer 

 #1
avatar+1493 
+4

If I am not mistaken, this problem requires you to generate the cubic equation of the four points listed. I see that \(f(0)=0\), so one of the coordinates of the x-intercept is certainly located at \((0,0)\).I also notice that there must be another zero located at \(1  because the graph crosses the x-axis there, but we cannot determine exactly where it is yet. An odd-degree polynomial cannot have an even number of real zeros (unless a multiplicity of a zero is even). We have determined already that there are 2 zeros, so a third must be lying somewhere. Yet again, we cannot determine where. 

 

Cubic functions are written in the form \(f(x)=ax^3+bx^2+cx+d\). We can actually solve for d straightaway by plugging in \(f(0)\).

 

\(f(x)=ax^3+bx^2+cx+d\)Plug in \(f(0)\)
\(f(0)=a*0^3+b*0^2+c*0+d\)There is a lot of simplification that can occur here!
\(f(0)=d\)According to your original three points, \(f(0)=0\), so substitute that in.
\(0=d\) 
  

 

Great! We have already solved for one of the missing variables. Our original equation is now \(f(x)=ax^3+bx^2+cx\). It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!

 

\(f(x)=ax^3+bx^2+cx\)Evaluate this function at f(-1).
\(f(-1)=a(-1)^3+b(-1)^2+c(-1)\)Ok, now simplify.
\(f(-1)=-a+b-c\)By definition, \(f(-1)=15\).
\(15=-a+b-c\) 
  

 

Let's plug in the second point.

 

\(f(1)=a(1)^3+b(1)^2+c(1)\)Simplify from here.
\(f(1)=a+b+c\)\(f(1)=-5\), so plug that back in!
\(-5=a+b+c\) 
 

 

 

Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique. 

 

\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)

 

Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.

 

\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)Add the equations together.
\(10=2b\)Divide by 2 on both sides!
\(5=b\) 
  

 

Now, let's plug in the third point.

 

\(f(x)=ax^3+bx^2+cx\)Evaluate the function at \(f(2)\)
\(f(2)=a(2)^3+b(2)^2+c(2)\)Now, simplify.
\(f(2)=8a+4b+2c\)Remember that \(f(2)=12 \) and \(b=5\).
\(12=8a+20+2c\)Subtract 20 from both sides.
\(-8=8a+2c\)Divide by 2 because that is the GCF of the coefficients, and it will make the numbers nicer in the long run.
\(-4=4a+c\) 
  

 

Ok, let's compare the previous equation with the first one we calculated. 

 

\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)

 

It looks like more cancellation can be done, right? Well' let's do that then!

 

\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)Add both equations together.
\(11=3a+b\)Of course, \(b=5\).
\(11=3a+5\)Now, solve for a.
\(6=3a\) 
\(2=a\) 
  

 

Now, plug this back into an already existing equation to find c.

 

\(-4=4a+c\)\(a=2\), so substitute that value in.
\(-4=4*2+c\)Now, solve for c.
\(-4=8+c\)Subtract 8 from both sides and isolate c.
\(-12=c\) 
  

 

Ok, our function is now clearly defined for the points given. It is now \(f(x)=2x^3+5x^2-12x\). It is time to find the remaining zeros!

 

\(2x^3+5x^2-12x=0\)Factor out a GCF first before doing anything else!
\(x(2x^2+5x-12)=0\)The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions.
\(x(2x^2-3x+8x-12)=0\)Now, factor by grouping. 
\(x[x(2x-3)+4(2x-3)]\) 
\(x(x+4)(2x-3)=0\)Now, set each factor equal to zero and solve.
\(x=0\)\(x+4=0\\ x=-4\)\(2x-3=0\\ 2x=3\\ x=\frac{3}{2}=1.5\)

 

All the zeros have been solved for finally.
  

 

Therefore, the zeros are \(-4,0,1.5\).

TheXSquaredFactor  Nov 29, 2017
edited by TheXSquaredFactor  Nov 29, 2017
Sort: 

2+0 Answers

 #1
avatar+1493 
+4
Best Answer

If I am not mistaken, this problem requires you to generate the cubic equation of the four points listed. I see that \(f(0)=0\), so one of the coordinates of the x-intercept is certainly located at \((0,0)\).I also notice that there must be another zero located at \(1  because the graph crosses the x-axis there, but we cannot determine exactly where it is yet. An odd-degree polynomial cannot have an even number of real zeros (unless a multiplicity of a zero is even). We have determined already that there are 2 zeros, so a third must be lying somewhere. Yet again, we cannot determine where. 

 

Cubic functions are written in the form \(f(x)=ax^3+bx^2+cx+d\). We can actually solve for d straightaway by plugging in \(f(0)\).

 

\(f(x)=ax^3+bx^2+cx+d\)Plug in \(f(0)\)
\(f(0)=a*0^3+b*0^2+c*0+d\)There is a lot of simplification that can occur here!
\(f(0)=d\)According to your original three points, \(f(0)=0\), so substitute that in.
\(0=d\) 
  

 

Great! We have already solved for one of the missing variables. Our original equation is now \(f(x)=ax^3+bx^2+cx\). It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!

 

\(f(x)=ax^3+bx^2+cx\)Evaluate this function at f(-1).
\(f(-1)=a(-1)^3+b(-1)^2+c(-1)\)Ok, now simplify.
\(f(-1)=-a+b-c\)By definition, \(f(-1)=15\).
\(15=-a+b-c\) 
  

 

Let's plug in the second point.

 

\(f(1)=a(1)^3+b(1)^2+c(1)\)Simplify from here.
\(f(1)=a+b+c\)\(f(1)=-5\), so plug that back in!
\(-5=a+b+c\) 
 

 

 

Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique. 

 

\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)

 

Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.

 

\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)Add the equations together.
\(10=2b\)Divide by 2 on both sides!
\(5=b\) 
  

 

Now, let's plug in the third point.

 

\(f(x)=ax^3+bx^2+cx\)Evaluate the function at \(f(2)\)
\(f(2)=a(2)^3+b(2)^2+c(2)\)Now, simplify.
\(f(2)=8a+4b+2c\)Remember that \(f(2)=12 \) and \(b=5\).
\(12=8a+20+2c\)Subtract 20 from both sides.
\(-8=8a+2c\)Divide by 2 because that is the GCF of the coefficients, and it will make the numbers nicer in the long run.
\(-4=4a+c\) 
  

 

Ok, let's compare the previous equation with the first one we calculated. 

 

\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)

 

It looks like more cancellation can be done, right? Well' let's do that then!

 

\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)Add both equations together.
\(11=3a+b\)Of course, \(b=5\).
\(11=3a+5\)Now, solve for a.
\(6=3a\) 
\(2=a\) 
  

 

Now, plug this back into an already existing equation to find c.

 

\(-4=4a+c\)\(a=2\), so substitute that value in.
\(-4=4*2+c\)Now, solve for c.
\(-4=8+c\)Subtract 8 from both sides and isolate c.
\(-12=c\) 
  

 

Ok, our function is now clearly defined for the points given. It is now \(f(x)=2x^3+5x^2-12x\). It is time to find the remaining zeros!

 

\(2x^3+5x^2-12x=0\)Factor out a GCF first before doing anything else!
\(x(2x^2+5x-12)=0\)The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions.
\(x(2x^2-3x+8x-12)=0\)Now, factor by grouping. 
\(x[x(2x-3)+4(2x-3)]\) 
\(x(x+4)(2x-3)=0\)Now, set each factor equal to zero and solve.
\(x=0\)\(x+4=0\\ x=-4\)\(2x-3=0\\ 2x=3\\ x=\frac{3}{2}=1.5\)

 

All the zeros have been solved for finally.
  

 

Therefore, the zeros are \(-4,0,1.5\).

TheXSquaredFactor  Nov 29, 2017
edited by TheXSquaredFactor  Nov 29, 2017
 #2
avatar+278 
+1

Thank you so much!  You're a life saver!

AnonymousConfusedGuy  Nov 30, 2017

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