Quadratic Inequalities in Two Variables

hey guys, this is ashley, i was supposed to blog yesterday but forgot so here goes...

Quadratic inequalities in two variables can be expressed in one of the four forms:

y < ax² + bx + c when either of these to inequalities is used you will graph
y < ax² + bx + c with a dotted line...
y ≤ ax ² + bx +c with either of these two inequalities, you will graph with
y ≥ ax² + bx + c a solid line





where a, b, and c are real numbers and a does not equal to zero

• A quadratic inequality in two variables represents a region on a cartesian plain with a parabola as the boudary. The graph of a quadratic inequality is the set of points (x,y) which are solutions to the inequality.

Example 1.

graph y< -2 (x - 3)² + 1. Is the point (2, -4) an inequality?

We know: because we have < in our inequality, we know that we are going to graph with a dotted line. that the x value serves as the axis of symmetry, and that the parabola will open downwards.

1. start with finding the vertex which is at (3, 1)
   

2. pick a point such as x=4. we have to find the y value and we will go about doing this by plugging the x value into the equation. y< -2 (x - 3)² + 1..... y< -2 (4-3)² +1 ..... y=-2(1)² +1 ...... y= -2 +1 ..... y=-1 Now we can reflect this point over the access of symmetry (2, -1)
3. find which area to shade. to do this you need two test points, one inside the parabola, and one outside the parabloa. you can pick any points, we will use (3, -3) inside, and (0,0) for the outside points. now just solve and find which is true and false. we will start with the inside points: y< -2 (x-3)² +1 .... -3 < -2(3-3)² +1 ... -3 < -2(0) +1 .... -3 < 1 which is true which means we will shade the inside of the parabola. to further prove this we can solve for the outside points and see if they are false: y<-2(x-3)+1 ... 0 < -2(0-3)²+1 ... 0 < -2 (9) +1... 0<-18+1 ... 0<-17 which is false.

and that is how you would go about solving this quadratic inequality with two variables.