# parabola equation

• June 5th 2010, 06:35 PM
rosana
parabola equation
Hi everyone,
I have a simple question but I can't solve it
what is the intersection point between these two parabola
$y=x^2$
and
$x=4y-y^2$

Thanks for any kind of help.
• June 5th 2010, 06:38 PM
11rdc11
Quote:

Originally Posted by rosana
Hi everyone,
I have a simple question but I can't solve it
what is the intersection point between these two parabola
$y=x^2$
and
$x=4y-y^2$

Thanks for any kind of help.

$y=(4y-y^2)^2$

Now expand and solve
• June 5th 2010, 07:17 PM
rosana
when I expand the equation $y=(4y-y^2)^2$
it will be
$y^4-8y^3+16y^2-y=0$
I know that $y=0$
but what about the other solution of the equation?
• June 5th 2010, 07:51 PM
11rdc11
Quote:

Originally Posted by rosana
when I expand the equation $y=(4y-y^2)^2$
it will be
$y^4-8y^3+16y^2-y=0$
I know that $y=0$
but what about the other solution of the equation?

Use the Intermediate Value Theorem
• June 5th 2010, 08:53 PM
rosana
I know the meaning of intermediate value theorem is that there is apoint c in interval $[a,b]$ for a function f verifying
$f'(c)=(f(b)-f(a))/(b-a)$

How is this related to the solution of the equation?
• June 5th 2010, 09:21 PM
11rdc11
Quote:

Originally Posted by rosana
I know the meaning of intermediate value theorem is that there is apoint c in interval $[a,b]$ for a function f verifying
$f'(c)=(f(b)-f(a))/(b-a)$

How is this related to the solution of the equation?

College Algebra Tutorial on Zeros of Polynomial Functions, Part II