solving inequalities

Question

What is the inequality for y<x^2-2x+3

Answer 1

Here given only one inequality is

y <x^2-2x+3

Now i factorize x^2-2x+3

x = -b±√(b^2-4ac)/2a

x = 2±√(4-12)/2

x = 2±√-8/2

x = 2±√(4*-2)/2

x = 2±2√-2/2

x = 2(1±√2i)/2

x = 1±√2i

y < (1+√2i)(1-√2i)

Answer 2

The inequality is y < x² - 2x + 3.

Write the equation y = x² - 2x + 3 and it represents a parabola curve.

The graph of the inequality y < x² - 2x + 3 is the shaded region, so every point in the shaded region satisfies the inequality.

The graph of the equation y = x² - 2x + 3 is the boundary of the region. Since the inequality symbol is <, the boundary is drawn as a dotted curve to show that points on the curve doesnot satisfy the inequality.

To graph the boundary curve make the table.

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for x and find the corresponding values for y.

x	y = x2 - 2x + 3	(x, y)
- 2	y = (-2)2 - 2(-2) + 3 = 4 + 4 + 3 = 11	(- 2, 11)
- 1	y = (-1)2 - 2(-1) + 3 = 1 + 2 + 3 = 6	(- 1, 6)
0	y = (0)2 - 2(0) + 3 = 0 + 0 + 3 = 3	(0, 3)
1	y = (1)2 - 2(1) + 3 = 1 - 2 + 3 = 2	(1, 2)
2	y = (2)2 - 2(2) + 3 = 4 - 4 + 3 = 3	(2, 3)
3	y = (3)2 - 2(3) + 3 = 9 - 6 + 3 = 6	(3, 6)
4	y = (4)2 - 2(4) + 3 = 16 - 8 + 3 = 11	(4, 11)

To draw inequality y < x² - 2x + 3 follow the steps.

1.  Draw a coordinate plane.

2.  Plot the points and draw a smooth curve through these points.

3.  To determine which side (out side or in side) to be shaded, use a test point inside the parabola. A simple choice is (1, 6).

Substitute the value of (x, y) = (1, 6) in the original inequality.

6 < 1² - 2(1) + 3

6 < 1 - 2 +3

6 < 2.

4.  Since the above statement is false, shade the region outside the parabola.