# Finding the point of intersection between a linear and quadratic function.

$$x-2y=2$$ $$y^2-x^2=2x+4$$

Find the points of intersection between these two functions.

I haven't done these problems before but I tried to use substitution (treating it like two linear functions, which I have done before).

$$x=2+2y\text{ , } y^2-x^2=2x+4$$ Substituting $x$ into the quadratic function gives: $$y^2-(2+2y)^2=2(2+2y)+4$$ $$-3y^2-12y-12=0$$

I am not sure if this is the correct approach to finding the points of intersection. If so, I don't know the next step. Could someone please explain? Thank you!

*Edit:

Is $-3y^2-12y-12=0$ equivalent to $3y^2+12y+12$? I just moved each term to the other side of the equation...

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Looks like a good start to me!

From

$$y^2 + 4y + 4 = 0$$

you can solve for $y$ perhaps by inspection, or by using the quadratic formula:

$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2} = -2.$$

Then, $x = 2 + 2y = 2 + 2(-2) = -2.$

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I redid my algebra and noticed I made an error, but I found the point of intersection to be $(-2,-2)$. –  Kot Dec 12 '13 at 4:45
Yep, I think that's right. There's also the possibility with other problems that you can have two points of intersection, or no points of intersection, based on whether the part underneath the square root sign is positive or negative, respectively. –  John Dec 12 '13 at 4:50
This problem has only one intersection because the part under the root is zero, correct? –  Kot Dec 12 '13 at 4:54
That's exactly right. –  John Dec 12 '13 at 4:54

Substituting $x$ into the quadratic function gives: \begin{align} y^2 - (2+2y)^2 &= 2(2+2y)+4\\ 3y^2 + 12y + 12 & =0\\ y^2+4y + 4 &= 0 \end{align}

So we obtain $y=-2, x= 2+2y=-2$, $(x,y)=(-2,-2)$, which satisfy the original equations.

From the point of the geometry, the first equation represents a line, and the second represents a hyperbola with center at $(-1,0)$. There is only one point that the two intersect. The point is $(-2,-2)$.

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