Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,708 users

Graph the following inequality on the graph provided.

0 votes

y<-(x-1)^{2}-2

asked Aug 7, 2014 in ALGEBRA 2 by anonymous

1 Answer

0 votes

The quadratic inequality y < - (x - 1)2 - 2

y < - (x2 + 1 - 2x) - 2

y < - x2 + 2x - 3

It's related equation y = - x2 + 2x - 3

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)

- 1

y = - (-1)2 + 2(-1) - 3

(- 1, -6)
-2

y = - (-2)2 + 2(-2) - 3

(- 2, -11)

0

y = - (0)2 + 2(0) - 3

(0, -3)

3

y = - (3)2 + 2(3) - 3

(3, -6)

4

y = - (4)2 + 2(4) - 3

(4, -11)

-3

y = - (-3)2 + 2(-3) - 3

(-3, -18)

              5

y = - (5)2 + 2(5) - 3

(5, -18)

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, -4).

Substitute the value of (x, y) = (1, -4) in the original inequality.

-4 < -(1-1) 2 - 2

-4 < -2

4.  Since the above statement is true, shade the region inside the parabola.

 

answered Aug 7, 2014 by david Expert

Related questions

asked Dec 30, 2018 in ALGEBRA 1 by anonymous
asked Sep 25, 2014 in PRECALCULUS by anonymous
...