How Do You Form and Expand a Cubic Equation from Given Intersections?

[Physiona]

Shown below is the graph of

The graph interests the

axis at 3 points, B, C and D.

Given the points of intersection, and the brackets below, form and expand an equation for

the graph of

ATTEMPT:
I've assumed to involve the points of intersection for the x-axis which is 4, -2, and -5. It looks to be a cubic graph where x³. I'm not entirely convinced on how I progress from here if this way is correct. Any guidance? Thank you.

 

[tnich]

Shown below is the graph of View attachment 222920

The graph interests the View attachment 222921axis at 3 points, B, C and D.

Given the points of intersection, and the brackets below, form and expand an equation for

the graph of View attachment 222922View attachment 222923


View attachment 222924

ATTEMPT:
I've assumed to involve the points of intersection for the x-axis which is 4, -2, and -5. It looks to be a cubic graph where x³. I'm not entirely convinced on how I progress from here if this way is correct. Any guidance? Thank you.

When you factor a cubic, you get something that looks like $(x-a)(x-b)(x-c)$. What do a, b and c represent?

 

[fresh_42]

Shown below is the graph of View attachment 222920

The graph interests the View attachment 222921axis at 3 points, B, C and D.

Given the points of intersection, and the brackets below, form and expand an equation for

the graph of View attachment 222922View attachment 222923


View attachment 222924

ATTEMPT:
I've assumed to involve the points of intersection for the x-axis which is 4, -2, and -5. It looks to be a cubic graph where x³. I'm not entirely convinced on how I progress from here if this way is correct. Any guidance? Thank you.

What does it mean for a polynomial $f(x)$ that $f(a)=0\,$? What happens, if you divide such a polynomial by $x-a\,$?

 

[Physiona]

When you factor a cubic, you get something that looks like $(x-a)(x-b)(x-c)$. What do a, b and c represent?

Will they not represent the points of intersection on the graph on the x axis? I'm not entirely sure

 

[Physiona]

What does it mean for a polynomial $f(x)$ that $f(a)=0\,$? What happens, if you divide such a polynomial by $x-a\,$?

I don't understand your point clearly. At this stage I don't have the value of the equation of the function.

 

[tnich]

Will they not represent the points of intersection on the graph on the x axis? I'm not entirely sure

That's right. They are the roots of the equation, the values of x at which f(x) = 0. So if f(x) = (x-a)(x-b)(x-c), then . . . ?

 

[fresh_42]

I don't understand your point clearly. At this stage I don't have the value of the equation of the function.

Polynomials can be divided. This means for two polynomials $f(x)$ and $g(x)$ we can find polynomials $q(x)$ and $r(x)$ such that $f(x)=q(x)\cdot g(x) + r(x)$ with $\deg r(x) < \deg g(x)$. Now insert $a$ in this equation with $f(a)=0$ and a suitable $g(x)$.

 

[Physiona]

That's right. They are the roots of the equation, the values of x at which f(x) = 0. So if f(x) = (x-a)(x-b)(x-c), then . . . ?

Will $(x-a)$= $(x-5)$ and $(x-b)$= $(x-2)$ and then $(x-c)$= $(x+4)$
And then factorised out to form an equation is x³+4x²-2x²-8x-5x²-20x?

 

[tnich]

Will $(x-a)$= $(x-5)$ and $(x-b)$= $(x-2)$ and then $(x-c)$= $(x+4)$
And then factorised out to form an equation is x³+4x²-2x²-8x-5x²-20x?

Not quite. You have a sign wrong in one of the factors.

 

[Physiona]

Not quite. You have a sign wrong in one of the factors.

Is $(x+4)$, $(x-4)$? Is that the error?

 

[tnich]

Is $(x+4)$, $(x-4)$? Is that the error?

Nope.

 

[tnich]

Nope.

That is one error, but you have similar errors in the other factors. If a = -2, what is (x-a)?

 

[Physiona]

That is one error, but you have similar errors in the other factors. If a = -2, what is (x-a)?

Oh, hang on. Is it $(x+2)$ (because (x-(-2)) makes x+2..
And $(x+5)$, and then $(x-4)$?
Is this right?

 

[tnich]

Oh, hang on. Is it $(x+2)$ (because (x-(-2)) makes x+2..
And $(x+5)$, and then $(x-4)$?
Is this right?

Yes, that's right. Think about what it means to say f(x) = (x-a)(x-b)(x-c). If x = a, what can you say about f(x)?

 

[Ray Vickson]

Oh, hang on. Is it $(x+2)$ (because (x-(-2)) makes x+2..
And $(x+5)$, and then $(x-4)$?
Is this right?

You tell us!

 

[Physiona]

Will $f(x)$ be $f(-2)$? as under this circumstance, $a=-2$..

 

[tnich]

Will $f(x)$ be $f(-2)$? as under this circumstance, $a=-2$..

That's true, but not what I was getting at. If f(x) = (x-a)(x-b)(x-c) and you substitute a for x in the equation, what answer do you get?

 

[Physiona]

That's true, but not what I was getting at. If f(x) = (x-a)(x-b)(x-c) and you substitute a for x in the equation, what answer do you get?

Will I have to substitute $a=-2$ in all of the polynomial do for x?
I.e. $(x-b)$ ---> $(-2+5)$?
And then multiply them all out? Please correct me if I'm wrong as I'm not sure.

 

[tnich]

Will I have to substitute $a=-2$ in all of the polynomial do for x?
I.e. $(x-b)$ ---> $(-2+5)$?
And then multiply them all out? Please correct me if I'm wrong as I'm not sure.

That is the basic idea, but try it and I think you will quickly see a shortcut.

 

[LCKurtz]

And when you get your arithmetic all settled out, don't forget you are also given $f(0) = -40$ in the picture.

 

[Physiona]

That is the basic idea, but try it and I think you will quickly see a shortcut.

I get $(-2+2)(-2+5)(-2-4)$
When I multiply it out I get zero for one set ---> $(-2+2)(-2+5)$ =0.
They all cancel out to zero. Meaning $x$ is zero..?
P.S: does $y= f(x)$ represent the y intercept by any chance? So $f(0) = -40$?

 

[Physiona]

The final equation I get is x³+3x²-18x-40

 

[tnich]

You started with f(x) = (x-a)(x-b)(x-c). Then you set x = a, so you substituted a for x in the equation to give you f(a) = (a-a)(a-b)(a-c) = 0. You can see that if you set x = a or b or c, you will get the same result. The point is, if you know what values make a polynomial equal to zero (the roots), you can write the polynomial.

 

[tnich]

The final equation I get is ##x³+3x²-18x-40##

That's right.

 

[Physiona]

You started with f(x) = (x-a)(x-b)(x-c). Then you set x = a, so you substituted a for x in the equation to give you f(a) = (a-a)(a-b)(a-c) = 0. You can see that if you set x = a or b or c, you will get the same result. The point is, if you know what values make a polynomial equal to zero (the roots), you can write the polynomial.

Right that makes sense now. Thank you.

 

[Physiona]

That's right.

I have another question if that's okay to ask? It's not based on the same topic though..

 

[LCKurtz]

You started with f(x) = (x-a)(x-b)(x-c). Then you set x = a, so you substituted a for x in the equation to give you f(a) = (a-a)(a-b)(a-c) = 0. You can see that if you set x = a or b or c, you will get the same result. The point is, if you know what values make a polynomial equal to zero (the roots), you can write the polynomial.

He should have started with $k(x-a)(x-b)(x-c)$ with $k$ unknown. It was just luck he got $f(0)=-40$.

 

[LCKurtz]

I have another question if that's okay to ask? It's not based on the same topic though..

No. Start a new topic.

 

[Physiona]

He should have started with $k(x-a)(x-b)(x-c)$ with $k$ unknown. It was just luck he got $f(0)=-40$.

Erm, are you talking about me? I'm a female and I've finished the problem thank you very much. I know where I went wrong and my mistakes, and yes I did intend to do that method at the start just didn't post it on here as I was looking for an alternative to quickly approach if, so it wasn't luck. Thank you either way.

 

[LCKurtz]

You're welcome. Sorry, hard to tell gender on these boards. Truth is, I don't know whether Physonia is a proper name in the first place, or, if it is, whether male or female.

 

[Mark44]

Thread moved, as this is really a precalculus type question.
Also, @Physiona, please stop deleting the homework template.