Question

Can somebody explain how the solutions manual got the attached graph, for the system
2x -y > 4
y< -(x*x) + 2?
I came up with the below table..
x =1, 0, -1, 2, -2
y = 1, 2, 1, -2, -2
I keep wanting to draw my parabola upwards instead of downwards, like the solutions manual does it...and I'm trying to follow my table closely when I do so...can somebody maybe figure out where I'm going wrong?

Answer 1

Somebody please help!

Answer 2

Hello again..

You can see that the parabola will open downward because the coefficient on the \(x^2 \) term is negative.

Answer 3

Basically you again want to start your graphs by drawing the curves that represent the edges of your inequalities.

Answer 4

So how do you explain the positive 1 value for x, and the positive corresponding value of 1 for y? How am I to graph that point?

Answer 5

So graph the lines:

\[y = -x^2 + 2\]and\[2x - y = 4\]

Answer 6

I have the 2x -y = 4 graphed already...

Answer 7

When dealing with inequalities, always start by graphing the equality first. Then you just have to figure out where to shade.

Answer 8

Excellent.

Answer 9

So graph
\[y = -x^2 +2\]

Answer 10

If I was to put 1 in for the value of x , that would make y have a value of 1 right?

Answer 11

That's correct.

I think some of the other values in your table are incorrect however.

\[-(-1)^2 + 2 = -1 + 2 =1\]

Answer 12

what values in my table are incorrect? And how am I to graph the point 1,1...without it making my parabola go upwards?

Answer 13

At 0, your parabola is a 2. At \(\pm 1\) your parabola is at 1. At \(\pm 2\) your parabola is at -2

Answer 14

As you move farther and farther away from 0, your graph decreases further and further. Hence opening downward.

Answer 15

so the y values for -1 and -2 are wrong?

Answer 16

Indeed.

Answer 17

\[-(-2)^2+2 = -4 + 2 = -2\]

Answer 18

I'm hitting a mental block here...I thought that I had when -2 equals x, y = -2...which would answer your above equation, right?

Answer 19

Sorry, yes your values are correct. I misread them.

I'm not sure why (looking at those values) you think that the function is not decreasing as you move away from x=0.

Answer 20

Graph it starting at \(x=0\implies y=2\). Then move right 1 unit and you have \(x=1\implies y=1\). As you pick bigger and bigger numbers your curve will continue to go down. Same as you move to the left from 0.

Answer 21

I'm caught up on 1,1...if I start at x =0, y = 2 (or the y intercept of 2, right?)...I'd go up one then one to the right?

Answer 22

You have y=2. Then you have y=1. You went down one, not up.

Answer 23

As your x moves to the right, your y moves down. As your x moves to the left your y moves down. The highest point on the curve is y=2 when x = 0.

Answer 24

I'm starting to spark..but the light bulb is not on yet..:P

Answer 25

So if I start at the y intercept...for 1,1 I'd go down 1, then to the right one..?

Answer 26

Draw the point (0,2). Then draw the point (1,1) Then draw the point (-1,1)

Answer 27

Then connect the dots =)

Answer 28

for other equations....where the x squared is positive..it's addition to the y intercept...and when the x squared is negative..it's a subtraction of the y intercept, in a sense?

Answer 29

You can think about it that way, but really it's just easier to draw in your points and connect the dots.

Answer 30

I need to think about it in some way....otherwise...whenever I see a positive number for both x and y in my table..I'm going to want to go up...

Answer 31

No you won't. The best way to think about it is to draw your dots. Trust me.

Answer 32

I can't draw dots...unless I know how to interpret my table...

Answer 33

Certainly you can. The point (0,2) is the intersection of the lines y=2 and x=0.

Answer 34

The point (1,1) is the intersection of the lines y=1 and x=1.

Answer 35

Does that make sense?

Answer 36

When you make a table you are making a list of points that are on your curve.

Answer 37

I think it's making sense now....I'll see here soon...because I have another problem involving a parabola..

Answer 38

Just plug in a value for x, get a value for y, then take the pair of them and draw that point.

Answer 39

I hope that helps. I'm going to go have dinner. I'll talk to you later or you can email me if you like.