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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?

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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!

Nov 28, 2017

#1
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+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 $$-4,0,1.5$$.

Nov 29, 2017
edited by TheXSquaredFactor  Nov 29, 2017

#1
+2324
+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
#2
+1442
+2

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

AnonymousConfusedGuy  Nov 30, 2017