I made a math qustion to see your guys reaction.
y=x^2*4x-5
x=2y^2*5y-59
Hi Kallamoar,
\(y=x^2*4x-5\\ y=4x^3-5\)
\(x=2y^2*5y-59\\ x=10y^3-59\)
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I do not know what youare after but here is how you can map from one to the other
\(y=4x^3-5 \)
multiply all the y vaues by 2.5. that means it gets vertically stretched by a factor of 2.5
\(y=2.5(4x^3-5)\)
Now drop the whole graph by 46.5 units
\(y=2.5(4x^3-5) - 46.5\)
Now simplify the last equation
\(y=10x^3-59\)
Now reflect this last graph in the line y=x
and you get
\(x=10y^3-59\)
And that is how you can transform the graph \(y=4x^3-5\qquad into \qquad x=10y^3-59\)
Cool - don't you think.
Here is the interactive graph. You can play with it if you want :))
I made a math qustion to see your guys reaction.
y=x^2*4x-5
x=2y^2*5y-59
x and y have many roots, most of them "complex". Here are just two "real" roots:
x=1.1946 and y=1.81908
Hi Kallamoar,
\(y=x^2*4x-5\\ y=4x^3-5\)
\(x=2y^2*5y-59\\ x=10y^3-59\)
-------------------------------------------
I do not know what youare after but here is how you can map from one to the other
\(y=4x^3-5 \)
multiply all the y vaues by 2.5. that means it gets vertically stretched by a factor of 2.5
\(y=2.5(4x^3-5)\)
Now drop the whole graph by 46.5 units
\(y=2.5(4x^3-5) - 46.5\)
Now simplify the last equation
\(y=10x^3-59\)
Now reflect this last graph in the line y=x
and you get
\(x=10y^3-59\)
And that is how you can transform the graph \(y=4x^3-5\qquad into \qquad x=10y^3-59\)
Cool - don't you think.
Here is the interactive graph. You can play with it if you want :))