Question

The shape of a roller coaster is modeled by a polynomial function, R(x). Describe how to find the x-intercepts of R(x) and how to construct a rough graph of R(x) so that the engineer can predict when there will be no change in the direction of the coaster. You may create a sample polynomial to be used in your explanations.

Answer 1

@SolomonZelman

Answer 2

got disconnected

Answer 3

ok

Answer 4

For any \(\large\color{black}{ R(x)=..............}\) , to find the x-intercepts.
set: \(\large\color{black}{ 0=..............}\)

Answer 5

see what I am doing?

Answer 6

and then you solve for x, for whatever the function is.

Answer 7

no

Answer 8

explain please

Answer 9

yes, so for example if we have:
\(\large\color{black}{ R(x)=x^3-2x^2-x+2}\)
then you set: \(\large\color{black}{ 0=x^3-2x^2-x+2}\) ,and solve for x.

Answer 10

\(\large\color{black}{ 0=x^3-2x^2-x+2}\)
\(\large\color{black}{ 0=x^2(x-2)-(x-2)}\)
\(\large\color{black}{ 0=(x^2-1)(x-2)}\)
\(\large\color{black}{ 0=(x-1)(x+1)(x-2)}\)
so your x-intercepts would be (in my example) \(\large\color{black}{ 1,~-1,~~~2}\) .

Answer 11

so you would mark the points \(\large\color{black}{ (-1,0)}\) , \(\large\color{black}{ (1,0)}\) , \(\large\color{black}{ (2,0)}\) .

Answer 12

in your example I could usually imagine a very high degree function.
so you would have many x-intercepts, but this is ain't important

Answer 13

you are:
~ Setting R(x)=0 (replacing R(x) by a 0 )
~ Solve for x
~ Write the solutions as points)

Answer 14

ok let me try to understand

Answer 15

sure, take your time

Answer 16

R(x)=x3-2x2x+2
set R(x) to 0
0=x3-2x2-x+2
0=x2(x-4)(x-4)
0=(x2-2)(x-2)
0=(x-2)(x+2)(x-4)

Answer 17

@SolomonZelman is this correct so far?

Answer 18

were you doing, \(\large\color{black}{ R(x)=x^3-2x^2-x+2 }\) ?

Answer 19

you have a little error I think

Answer 20

yes i am and ok help me please

Answer 21

\(\large\color{black}{ R(x)=x^3-2x^2-x+2 }\)
\(\large\color{black}{ 0=x^3-2x^2-x+2 }\)
good till now,

then you factor:
\(\large\color{black}{ R(x)=x^3-2x^2-x+2 }\) \(\LARGE\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ 0=x^2(x-2)-(x-2) }\) \(\LARGE\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ 0=x^2(x-2)-1(x-2) }\) \(\LARGE\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ 0=(x^2-1)(x-2) }\) \(\LARGE\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ 0=(x^2-1^2)(x-2) }\) \(\LARGE\color{white}{ \rm \left| \right| }\)
\(\large\color{black}{ 0=(x+1)(x-1)(x-2) }\) the the zeros are: -1, 1, 2 \(\LARGE\color{white}{ \rm \left| \right| }\)

Answer 22

you don't have to use my function (that I offered as an example) though...

Answer 23

ok thanks i will post another one

Answer 24

sure.... (but remember, one question per thread)
yw, and good luck!

Answer 25

i know

Answer 26

god, yw 1s again

Answer 27

What was the answer ? ;-;