# Thread: Finding points of intersection

1. ## Finding points of intersection

Find the points of intersection of the graphs in the system.

x^2= 6y
y= -x

Please show me step by step how to solve this. The answers in the back of the book are

(0,0) and (-6,6).

EDIT:

Help also appreciated with the following problem:

x^2 + y^2 = 20
y= x-4

Find points of intersection.

2. All youve gotta do is solve for x or y and then substitute

watch

$\displaystyle x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}$

Now

since y=y

$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}$

How's that?

3. ## Points of intersection

Hello cuandoyando
Originally Posted by cuandoyando
Find the points of intersection of the graphs in the system.

x^2= 6y
y= -x

Please show me step by step how to solve this. The answers in the back of the book are

(0,0) and (-6,6).
A point on any graph is represented by a value of $\displaystyle x$ and a value of $\displaystyle y$ which are connected by the equation of the graph - in other words these values of $\displaystyle x$ and $\displaystyle y$ 'satisfy' the equation of the graph. We can find where two graphs intersect if we can find a value of $\displaystyle x$ and a value of $\displaystyle y$ which satisfy the equations of both graphs at the same time. The two equations you have are:

$\displaystyle x^2 = 6y$ and $\displaystyle y = -x$

So we need to find values of $\displaystyle x$ and $\displaystyle y$ that satisfy both these equations at the same time.

Now if $\displaystyle y = -x$, then $\displaystyle 6y = -6x$. So if, at the same time, $\displaystyle 6y = x^2$, then we must have:

$\displaystyle x^2 = -6x$

$\displaystyle \Rightarrow x^2+6x=0$

$\displaystyle \Rightarrow x(x+6) = 0$

$\displaystyle \Rightarrow x = 0$ or $\displaystyle -6$

So, bearing in mind that $\displaystyle y = -x$:

When $\displaystyle x = 0, y = 0$, and when $\displaystyle x = -6, y = 6$.

So the points where the graphs intersect are $\displaystyle (0,0)$ and $\displaystyle (-6,6)$.

Help also appreciated with the following problem:

x^2 + y^2 = 20
y= x-4

Find points of intersection.
Use the same technique here. If $\displaystyle y = (x-4)$, then we can replace $\displaystyle y$ by $\displaystyle (x-4)$ in the equation $\displaystyle x^2 + y^2=20$ (in other words, substitute for $\displaystyle y$) to get:

$\displaystyle x^2 + (x-4)^2 = 20$

Can you continue? (You'll need to expand the brackets, simplify and the solve the quadratic equation by formula.)

4. Originally Posted by VonNemo19
All youve gotta do is solve for x or y and then substitute

watch

$\displaystyle x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}$

Now

since y=y

$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}$

How's that?
I don't want to pick at you but you removed one solution by dividing by x:

$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2+6x=0}\Longleftrightarrow{ x(x+6)=0}$

A product equals zero if one factor equals zero. Therefore:

$\displaystyle x = 0~\vee~x=-6$