Cannot find the last intercept.. How?

**mitchelljk** · March 21st 2009, 07:07 PM

$(x^2 + y^2 -x)^2$ = $x^2 + y^2$

I found through doing some of the algebra that one x-intercept is (2,0) and that one y-intercept is (0,-1). And because that this equation is symmetical about the x-axis, another y-intercept is (0,1).

These intercepts are the only ones I could figure out on my own. The back of book says that the fourth intercept is (0,0). I cant figure out through my math, or through assumptions based on the equations symmerty why the remaining intercept is (0,0).

If anyone could explain to me how and why the remaining intercept is (0,0), that would be great! I've been trying to figure out why for about 2 hours now.

**Pinkk** · March 21st 2009, 07:37 PM

I know how to prove it using polar coordinates. Using the parameters $x=rcos\theta$ and $y=rsin\theta$ , then $x^{2}+y^{2}=r^{2}$ . Using these substitutions, we can rewrite the equation you gave as:

$(r^{2}-rcos\theta)^{2}=r^{2}$

Taking the square root of both sides gives you:

$r^{2}-rcos\theta=r$

Dividing through r gives you:

$r-cos\theta=1$

Which finally gives us:

$r=1+cos\theta$

Now, notice when $\theta=\pi$ , $r=0$ . When $r=0$ , the representation on the Cartesian coordinate system is the curve going through the origin. Knowing this, you can say that the remaining intercept is $(0,0)$ . To show algebraically that $(0,0)$ has to be an intercept, refer back to the equation $x^{2}+y^{2}=r^{2}$ . Plugging in $r=0$ , you get: $x^{2}+y^{2}=0^{2}$ . The solution to that is $x=0, y=0$ , or simply the point $(0,0)$ .

**mitchelljk** · March 22nd 2009, 02:29 AM

Sines Cosines and Tangents are a little over my head at this point. I dont understand why you decided to introduce $r^2$ to the $x^2 + y^2$ . It seems very random to me because I have never seen $r^2$ in a problem, other than if I recall correctly, circles. So could you please explain the reasoning behind the $r^2$ and where it comes from?

**The Second Solution** · March 22nd 2009, 02:41 AM

Originally Posted by mitchelljk

$(x^2 + y^2 -x)^2$ = $x^2 + y^2$

I found through doing some of the algebra that one x-intercept is (2,0) and that one y-intercept is (0,-1). And because that this equation is symmetical about the x-axis, another y-intercept is (0,1).

These intercepts are the only ones I could figure out on my own. The back of book says that the fourth intercept is (0,0). I cant figure out through my math, or through assumptions based on the equations symmerty why the remaining intercept is (0,0).

If anyone could explain to me how and why the remaining intercept is (0,0), that would be great! I've been trying to figure out why for about 2 hours now.

You know x = 0 is a solution because you got (0, 1) and (0, -1).

Substitute $x = 0$ : $y^4 = y^2 \Rightarrow y^4 - y^2 = 0 \Rightarrow y^2(y^2 -1) = 0 \Rightarrow y^2 (y - 1)(y + 1) = 0$ .

So when x = 0 then y = 0, 1, and -1.

**HallsofIvy** · March 22nd 2009, 04:16 AM

Originally Posted by mitchelljk

$(x^2 + y^2 -x)^2$ = $x^2 + y^2$

I found through doing some of the algebra that one x-intercept is (2,0) and that one y-intercept is (0,-1). And because that this equation is symmetical about the x-axis, another y-intercept is (0,1).

These intercepts are the only ones I could figure out on my own. The back of book says that the fourth intercept is (0,0). I cant figure out through my math, or through assumptions based on the equations symmerty why the remaining intercept is (0,0).

If anyone could explain to me how and why the remaining intercept is (0,0), that would be great! I've been trying to figure out why for about 2 hours now.

y intercepts occur where y= 0 so your equation becomes $(x^2- x)^2= x^2$ . Taking square roots of that, either $x^2- x= x$ or $x^2- x= -x$ . The first of those is the same as $x^2- 2x= x(x-2)= 0$ . Either x= 0 or x= 2. The second is the same as $x^2= 0$ so x= 0. In either case you get (0, 0) as a y-intercepts (and, of course, an x-intercept as well). I don't see why you didn't get that immediately.

**Pinkk** · March 22nd 2009, 07:00 AM

Originally Posted by mitchelljk

Sines Cosines and Tangents are a little over my head at this point. I dont understand why you decided to introduce $r^2$ to the $x^2 + y^2$ . It seems very random to me because I have never seen $r^2$ in a problem, other than if I recall correctly, circles. So could you please explain the reasoning behind the $r^2$ and where it comes from?

Well, the last two posts do a more direct evaluation, so I would just follow what they did. But if you're curious, remember $x=rcos\theta$ and $y=rsin\theta$ . Therefore, $x^{2}+y^{2}=r^{2}cos^{2}\theta + r^{2}sin^{2}\theta$ . Pulling the $r^{2}$ , we get $x^{2}+y^{2}=r^{2}(cos^{2}\theta + sin^{2}\theta)$ . One of the most fundamental trigonometric identities is $cos^{2}\theta+sin^{2}\theta=1$ , so substituting that into what we worked out, we get: $x^{2}+y^{2}=r^{2}(1)$ , or simply $x^{2}+y^{2}=r^{2}$ . Sorry if I mentioned material you haven't covered yet; in my high school, we were taught trigonometry and pre-calculus simutaneously.

**Soroban** · March 22nd 2009, 08:29 AM

Hello, mitchelljk!

I'm puzzled . . . How do you locate intercepts?

Find the intercepts: . $(x^2 + y^2 -x)^2 \:=\:x^2 + y^2$

To find $x$ -intercepts, let $y = 0$ and solve for $x.$

We have: . $(x^2 + 0^2 - x)^2 \:=\:x^2 + 0^2 \quad\Rightarrow\quad (x^2-x)^2 \:=\:x^2$

. . $x^4-2x^3 + x^2 \:=\:x^2 \quad\Rightarrow\quad x^4 - 2x^3 \:=\:0 \quad\Rightarrow\quad x^3(x-2) \:=\:0$

Hence: . $x \:=\:0,2 \quad\hdots\quad x\text{-intercepts: }\:(0,0),\:(2,0)$

To find $y$ -intercepts, let $x = 0$ and solve for $y.$

We have: . $(0^2 + y^2 - 0)^2 \:=\:0^2 + y^2 \quad\Rightarrow\quad y^4 - y^2 \:=\:0$

. . $y^2(y^2-1)\:=\:0 \quad\Rightarrow\quad y^2(y-1)(y+1)\:=\:0$

Hence: . $y \:=\:0,\:1,\:\text{-}1 \quad\hdots\quad y\text{-intercepts:} \:(0,0),\:(0,1),\:(0,\text{-}1)$

**mitchelljk** · March 24th 2009, 02:56 AM

Thanks everyone for your help. It makes perfect sense now. For some reason I never learned trig. So shortly I will be teaching it to myself.

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