need help with systems of inequalities?

Question

Here are the two equations:
y < 2x - 1
y >or= x^2 +3x - 7

I know how to do the first one-- m=2, y-intercept is (0, -1). But, how do I solve/graph the second equation???

Answer 1

My advice. Locate the equality line then put your pencil on it and note which side of the line the inequality is indicating. Shade that area, Shade the other inequality area differently. Then the area common to both inequalities solves both.

Answer 2

Note that the second equation is a parabola, with roots
x = [-3 +/- sqrt(9 + 28)]/2
x = [-3 +/- sqrt(37)]/2

So our x-intercepts are x = [-3 + sqrt(37)]/2 and [-3 - sqrt(37)]/2

What we have to do is note that there are two sides of a parabola; above and below. What you want to do is test x=0 and y=0; if you get a true value, you shade that side of the parabola (to represent the inequality). If you get a false value, you shade the other side of the parabola.

0 >= 0^2 + 3(0) - 7
0 >= -7 , which is a false statement.

So you shade below the parabola.

Answer 3

For the first one, you graphed the line y=2x-1 and shaded all the values below this line, that is y < this line.

For the second, it is a quadratic with a parabola for a graph. Shade the values above this graph, that is y >= the parabola.

For the first one y< line, we usually make a dotted line to indicate that it is not included. For the second y>=, we are including the graph of the parabola, so we make it a solid line.

To graph the parabola, find the zeros first, then the y-intercept to help you graph it.

Answer 4

First find the roots of the quadratic.

x = -3/2 +/- sqrt[3^2 + 4(1)(7)] /2

x = -3/2 +/- sqrt(37)/2

So we have

y >= [x + 3/2 + sqrt(37)/2][x + 3/2 - sqrt(37)/2]

This expression is a product. There are two ways it can be greater than or equal to 0. One way is for both terms to be nonnegative. The other way is for both terms to be nonpositive. Lets consider the first case.

x + 3/2 + sqrt(37)/2 >=0 AND x + 3/2 - sqrt(37)/2 >=0
x >= -3/2 - sqrt(37)/2 AND x >= -3/2 + sqrt(37)/2
Which means x >= -3/2 + sqrt(37)/2

The other solution is similar:

x + 3/2 + sqrt(37)/2 <=0 AND x + 3/2 - sqrt(37)/2 <=0
x <= -3/2 - sqrt(37)/2 AND x <= -3/2 + sqrt(37)/2
Which means x <= -3/2 - sqrt(37)/2

So the final solution is either x<= -3/2-sqrt(37)/2 (about -4.5) OR
x>=-3/2+sqrt(37)/2 (about 1.5)