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!
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. | |||
| 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 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. | |||
| All the zeros have been solved for finally. | |||
Therefore, the zeros are \(-4,0,1.5\).