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What are the coordinates of the points where f(x)=x^3-x^2+x+1 and g(x)=x^3+x^2+x-1 intersect?Asked by Trish Goal
What are the coordinates of the points where f(x) =x^3-x^2+x+1
And g(x)=x^3+x^2+x-1 intersect?
And g(x)=x^3+x^2+x-1 intersect?
Answers
Answered by
Susan
There is a program called Graph. You can google it by typing in "graph", download it for free, and put in those equations to find the intersection.
Answered by
Trish Goal
I did but I couldn't find the point of intersection.
Answered by
Susan
They intersect at two points. (1, 2) and (-1,-2).
Answered by
Reiny
you want g(x) = g(x) , so
x^3 + x^2 + x - 1 = x^3 - x^2 + x + 1
2x^2 = 2
x^2 = 1
x = ± 1
if x = 1 , g(1) = 1 - 1 + 1 + 1 = 2
if x = -1 , g(-1) = -1 - 1 - 1 + 1 = -2
so they intersect at (1,2) and (-1,-2)
x^3 + x^2 + x - 1 = x^3 - x^2 + x + 1
2x^2 = 2
x^2 = 1
x = ± 1
if x = 1 , g(1) = 1 - 1 + 1 + 1 = 2
if x = -1 , g(-1) = -1 - 1 - 1 + 1 = -2
so they intersect at (1,2) and (-1,-2)
Answered by
Susan
He means you want f(x) to equal g(x).
I.e. f(x) = g(x) [NOT g(x) = g(x)]
When you set two equations equal to each other you find the points of intersection.
I.e. f(x) = g(x) [NOT g(x) = g(x)]
When you set two equations equal to each other you find the points of intersection.
Answered by
Hasini
If the graphs of $f$ and $g$ intersect at $(x,y)$, then $y = f(x)=g(x)$, so
\[x^3+x^2+x-1=x^3-x^2+x+1.\]Moving all the terms to the left-hand side gives $2x^2-2=0$, so $x=\pm1$. Substituting into $g$ we find the points of intersection to be
\[
(-1,g(-1))=(-1,-2)
\text{~and~}
(1,g(1))=(1,2).
\]Therefore, our answer is $\boxed{(-1,-2), (1,2)}$.
\[x^3+x^2+x-1=x^3-x^2+x+1.\]Moving all the terms to the left-hand side gives $2x^2-2=0$, so $x=\pm1$. Substituting into $g$ we find the points of intersection to be
\[
(-1,g(-1))=(-1,-2)
\text{~and~}
(1,g(1))=(1,2).
\]Therefore, our answer is $\boxed{(-1,-2), (1,2)}$.
Answered by
YOMAma
Bro
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