points for y=x to the power of 1/3
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y = ∛x
Let's cube both sides.
This gives us,
y³ = x
now, let's find points to plot.
Assume y = 0
Then we know, 0³ = 0
Hence, for y = 0, x will be 0
So (0,0) is one point.
Now, assume y = 1
Then 1³ = 1 (So x = 1)
So, (1,1) is a valid solution
Assume y = 2
Then 2³ = 8 (So x = 8)
Then (2, 8) is a valid solution.
Similarly, we know that if we cube a negative number, we will get a negative cube. [just like -(2)³ = -8]
So the negative of all of the above points are also valid.
Whoaa! That gives us a lot of points:
(0, 0)
(2, 8)
[ (-2), (-8) ]
(1, 1)
[ (-1), (-1) ]
I hope that helps :)
Let's cube both sides.
This gives us,
y³ = x
now, let's find points to plot.
Assume y = 0
Then we know, 0³ = 0
Hence, for y = 0, x will be 0
So (0,0) is one point.
Now, assume y = 1
Then 1³ = 1 (So x = 1)
So, (1,1) is a valid solution
Assume y = 2
Then 2³ = 8 (So x = 8)
Then (2, 8) is a valid solution.
Similarly, we know that if we cube a negative number, we will get a negative cube. [just like -(2)³ = -8]
So the negative of all of the above points are also valid.
Whoaa! That gives us a lot of points:
(0, 0)
(2, 8)
[ (-2), (-8) ]
(1, 1)
[ (-1), (-1) ]
I hope that helps :)
aryan02p3jd6e:
Sorry, I mean (8, 2) and (-8, -2)
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