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 xintercepts of the graph of f?
Thank you all so much!
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 xintercept is certainly located at \((0,0)\).I also notice that there must be another zero located at \(1 because the graph crosses the xaxis there, but we cannot determine exactly where it is yet. An odddegree 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. 
\(f(1)=a+bc\)  By definition, \(f(1)=15\). 
\(15=a+bc\)  
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.
\(\hspace{2mm}15=a+bc\\ 5=+a+b+c\)
Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.
\(\hspace{2mm}15=a+bc\\ 5=+a+b+c\)  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.
\(\hspace{1mm}15=a+bc\\ 4=4a\hspace{7mm}+c\)
It looks like more cancellation can be done, right? Well' let's do that then!
\(\hspace{1mm}15=a+bc\\ 4=4a\hspace{7mm}+c\)  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 \(f(x)=2x^3+5x^212x\). It is time to find the remaining zeros!
\(2x^3+5x^212x=0\)  Factor out a GCF first before doing anything else!  
\(x(2x^2+5x12)=0\)  The next step involves finding factors of 24 that add up to 5. (3 and 8). Break up the linear term into these portions.  
\(x(2x^23x+8x12)=0\)  Now, factor by grouping.  
\(x[x(2x3)+4(2x3)]\)  
\(x(x+4)(2x3)=0\)  Now, set each factor equal to zero and solve.  
 All the zeros have been solved for finally.  
Therefore, the zeros are \(4,0,1.5\).
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 xintercept is certainly located at \((0,0)\).I also notice that there must be another zero located at \(1 because the graph crosses the xaxis there, but we cannot determine exactly where it is yet. An odddegree 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. 
\(f(1)=a+bc\)  By definition, \(f(1)=15\). 
\(15=a+bc\)  
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.
\(\hspace{2mm}15=a+bc\\ 5=+a+b+c\)
Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.
\(\hspace{2mm}15=a+bc\\ 5=+a+b+c\)  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.
\(\hspace{1mm}15=a+bc\\ 4=4a\hspace{7mm}+c\)
It looks like more cancellation can be done, right? Well' let's do that then!
\(\hspace{1mm}15=a+bc\\ 4=4a\hspace{7mm}+c\)  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 \(f(x)=2x^3+5x^212x\). It is time to find the remaining zeros!
\(2x^3+5x^212x=0\)  Factor out a GCF first before doing anything else!  
\(x(2x^2+5x12)=0\)  The next step involves finding factors of 24 that add up to 5. (3 and 8). Break up the linear term into these portions.  
\(x(2x^23x+8x12)=0\)  Now, factor by grouping.  
\(x[x(2x3)+4(2x3)]\)  
\(x(x+4)(2x3)=0\)  Now, set each factor equal to zero and solve.  
 All the zeros have been solved for finally.  
Therefore, the zeros are \(4,0,1.5\).