irst, here is the existing map of current structures. It is important that the rollercoaster does not go through the foundation of any of these structures. Pick three points on the x–axis that the rollercoaster will dive underground or resurface. (Points that have a building on them are off-limits.)
1.The rollercoaster will run from right to left, so work in that order when selecting points. •1st point:______
•2nd point:______
•3rd point:______

2.Using those points as zeros, construct the polynomial function, f(x), that will be the path of your rollercoaster. Show all of your work.

Question

irst, here is the existing map of current structures. It is important that the rollercoaster does not go through the foundation of any of these structures. Pick three points on the x–axis that the rollercoaster will dive underground or resurface. (Points that have a building on them are off-limits.)
1.The rollercoaster will run from right to left, so work in that order when selecting points. •1st point:______
•2nd point:______
•3rd point:______


 
2.Using those points as zeros, construct the polynomial function, f(x), that will be the path of your rollercoaster. Show all of your work.

anonymous · Answer

i only need help with number 2

anonymous · Answer

@phi @crazycatlady

anonymous · Answer

@.Sam.

anonymous · Answer

My points are (-8,0) (-3,0) and (6,0)

anonymous · Answer

@jdoe0001

amistre64 · Answer

you want a cubic .... with those roots ?

amistre64 · Answer

subtract x from all the x values, then multiply them all togther

f(x) = (-8-x)(-3-x)(6-x)

anonymous · Answer

I just need to know how to put it into a polynomial function using those points as zeros.

amistre64 · Answer

do you know how to multiply linear factors?

anonymous · Answer

so im confused. what is the x value? and No I don't think im familiar with linear factors. im in a lesson called solving polynomial equations

amistre64 · Answer

x is an unknown ... a polynomial equation is going to be constructed from a variable

amistre64 · Answer

a linear factor is just (x+a) for some real value a

amistre64 · Answer

does the term FOIL ring any bells?

anonymous · Answer

yes ive done FOIL.

amistre64 · Answer

FOIL is a method of multiplying 2 linear factors together ...

amistre64 · Answer

the real algebraic property at work is called "distribution"; but they dont like to teach actual math these days :/

anonymous · Answer

so am I doing distribution for this function or is the polynomial function gonna be f(x)=(-8-x)(-3-x)(6-x)?

amistre64 · Answer

(-8-x)(-3-x)(6-x)

lets multiply the first factors by distribution

(-8-x)(-3-x)

-8(-3-x) - x(-3-x)

-8(-3) -8(-x) - x(-3) -x(-x)

24 +8x +3x +x^2

24 +11x +x^2

yeah; then we distribute the third factor into that one

anonymous · Answer

ok thank you. that helps a lot.

amistre64 · Answer

the polymonial function has a specific definition ... it is defined to be a sum of terms

$$P(x)=c_o+c_1x+c_2x^2+c_3x^3+c_4x^4+....$$

amistre64 · Answer

since the factored form is not in the "polynomial form" then there might be someone that wants to complain ... but the are in fact equal representations of the same curve

anonymous · Answer

ok so when im multiplying (6-x), do I multiply the 6 by the 24 and then the x by 11x?

amistre64 · Answer

distrbuting in the last factor

(24 +11x +x^2) (6-x)

24(6-x) +11x (6-x) + x^2(6-x)

24(6) +24(-x) +11x(6) +11x(-x) + x^2(6) +x^2(-x)

144 -24x +66x -11x^2 + 6x^2 -x^3

144 +42x -5x^2 -x^3

anonymous · Answer

ok I understand that now. do I turn around the function so x^3 will be the first number or no?

amistre64 · Answer

it depends on how you want the curve to be presented.

amistre64 · Answer

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