# Label Corner Points

#### xyz_1965

Label corner points given the system of nonlinear inequalities.

#### topsquark

Forum Staff
By "corner points" I presume you are looking for intersection points on the boundary of the solution? These can be found by solving the simultaneous equations
$$\displaystyle x^2 - y = 0$$
$$\displaystyle 2x^2 + y = 12$$

So the first equation has the solution $$\displaystyle y = x^2$$. Put that into the second equation and solve for x. Can you finish?

-Dan

#### joshuaa

if you draw the two inequalities, you will see clearly where the corners are. you have to find the bounded area and then look for any intersections around it. to find any intersecting lines, first convert the inequalities to equal signs (=), then solve for y for both

y = x^2
y = 12 - 2x^2

then make them equal and solve for x

x^2 = 12 - 2x^2

here is a picture of the corners

#### xyz_1965

By "corner points" I presume you are looking for intersection points on the boundary of the solution? These can be found by solving the simultaneous equations
$$\displaystyle x^2 - y = 0$$
$$\displaystyle 2x^2 + y = 12$$

So the first equation has the solution $$\displaystyle y = x^2$$. Put that into the second equation and solve for x. Can you finish?

-Dan
Dan,

I can do it now. However, member joshuaa was nice enough to provide a complete solution and graph.

joshuaa

#### xyz_1965

if you draw the two inequalities, you will see clearly where the corners are. you have to find the bounded area and then look for any intersections around it. to find any intersecting lines, first convert the inequalities to equal signs (=), then solve for y for both

y = x^2
y = 12 - 2x^2

then make them equal and solve for x

x^2 = 12 - 2x^2

here is a picture of the corners

View attachment 39568
Nice reply and picture, too. Desmos, right?

joshuaa

#### joshuaa

Yes, Desmos. It is a great website to graph your functions!

xyz_1965