# Finding Horizontal Intercepts

• Jan 18th 2013, 11:34 PM
maca404
Finding Horizontal Intercepts
I have the function $x^4-2*x^2-1$

I have fount the dirivative
$4x^3-4x$

and all stationary points as the first part of the question asked but I am stuck on the second half it says :

(b) By using the substitution $u = x^2$, find the horizontal intercepts.

For the most part I dont really understand the question and have no idea were to start to solve it, Any ideas ?
• Jan 19th 2013, 12:21 AM
MarkFL
Re: Finding Horizontal Intercepts
I'm assuming by horizontal intercepts, they mean roots. Using the suggested substitution, we may write:

$f(u)=u^2-2u-1=0$

Now, use the quadratic formula (or complete the square) to find the roots in terms of $u$, then back substitute for $u$ to find the 4 roots in terms of $x$.
• Jan 19th 2013, 02:26 AM
maca404
Re: Finding Horizontal Intercepts
Yes its is the roots they are looking for and from graphing I know thay are -1.554 and 1.554, I can solve for the roots o the equation you suggested I just cant see how I am going to get back to the correct answer ?
• Jan 19th 2013, 03:33 PM
Prove It
Re: Finding Horizontal Intercepts
OK, first of all, do NOT round off until the very end. Keep answers exact where possible.

Now, as you were instructed by Mark, if you have let \displaystyle \begin{align*} u = x^2 \end{align*} then you end up with \displaystyle \begin{align*} u^2 - 2u - 1 = 0 \end{align*}. Solve this EXACTLY using the Quadratic Formula or Completing the Square. Then you will have the possible values for \displaystyle \begin{align*} u \end{align*}.

Once you have the values for \displaystyle \begin{align*} u \end{align*}, remember that \displaystyle \begin{align*} u = x^2 \end{align*} and use this to find the values of \displaystyle \begin{align*} x \end{align*}.
• Jan 19th 2013, 08:41 PM
maca404
Re: Finding Horizontal Intercepts
Ok using the quadratic equation I come up with

U= - $0.414213562$ and U = $2.414213562$

so if $U = X^2$ then $X=sqrt(u)$

and indeed $Sqrt(2.414213562)=1.554$

but I cant take the square of the neg so how do I get the value for -1.554 ?
• Jan 20th 2013, 07:23 PM
Prove It
Re: Finding Horizontal Intercepts
Quote:

Originally Posted by maca404
Ok using the quadratic equation I come up with

U= - $0.414213562$ and U = $2.414213562$

so if $U = X^2$ then $X=sqrt(u)$

and indeed $Sqrt(2.414213562)=1.554$

but I cant take the square of the neg so how do I get the value for -1.554 ?

• Jan 21st 2013, 12:23 AM
maca404
Re: Finding Horizontal Intercepts
You mean in the form x=1+sqrt(2) and x=1-sqrt*(2)
• Jan 21st 2013, 04:28 PM
Prove It
Re: Finding Horizontal Intercepts
Quote:

Originally Posted by maca404
You mean in the form x=1+sqrt(2) and x=1-sqrt*(2)

Almost. Actually it's \displaystyle \begin{align*} u = 1 \pm \sqrt{2} \end{align*}, not x.

Now since you know that \displaystyle \begin{align*} u = x^2 \end{align*}, that gives \displaystyle \begin{align*} x^2 = 1 \pm \sqrt{2} \end{align*}. How will you solve for x?