Tell whether the given points are on the graphs of the equations. State yes or no for each point. ?

Question

Equation: y= x^3 - 2√x

Points: (0,0), (1,1), (1, -1)

Equation: x^2 + y^2 = 4

Points: (0, 2), (-2, 2), (√2, √2)

Please explain - I want to understand how to do it. Thanks!

Answer 1

Just plug in each point.  If the equation holds true, then the point is on the graph.  If the equation doesn't hold true, then the point is not on the graph.

For example, (0,0) is on the graph y = x^3 - 2√x, because

0 = 0^3  - 2√0 = 0

But (1,1) is not, because plugging in those values gives

1 = 1^3 - 2√1

1 = 1 - 2

1 = -1

Answer 2

y= x³ - 2√x

x cannot be < 0, because √x would result in an imaginary number. So the line begins at (0,0). See diagram below:

http://i533.photobucket.com/albums/ee339...

x² + y² = 4

y² = 4 - x²

y = √(4 - x²)

x cannot be <-2 or >2, because √(4 - x²) would result in an imaginary number.

The line is a semicircle starting at (-2, 0), peaking at (0, 2) and ending at (2, 0). See image below:

http://i533.photobucket.com/albums/ee339...

Answer 3

(1,1) and (-2,2) are not on the graphs. all the others are

Answer 4

The first number in the brackets is x and the second y. Just substitute x and y for these numbers to see if the equation comes out right or not.

Answer 5

y= x^3 - 2√x

Let x = 0 -> y = 0 - 0 = 0 ... yes

Let x = 1 -> y = 1 - 2 = -1 ... no for (1,1), yes for (1,-1)

x^2 + y^2 = 4

For each point, plug in x and y values and see if the equation is satisfied.

(0,2) -> 0 + 4 = 4 ... yes

(-2,2) -> 4 + 4 = 8 ... no

(√2, √2) -> 2 + 2 = 4 ... yes

Answer 6

the first number in parenthesis is your x, and the second is your y. Plug them in, in the right places and if your answer works out then it is yes, if it is totally off (ex. 6=2) then it is no.