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OpenStudy (anonymous):
What equation is the inverse of y = 16x^2 + 1?
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OpenStudy (jhannybean):
Are you looking to find the inverse function of the one given?
OpenStudy (danjs):
Switch y and x around and solve for y
OpenStudy (anonymous):
Yes, @Jhannybean.
OpenStudy (jhannybean):
Ok, then I would start by what @DanJS has recommended as well.
OpenStudy (zzr0ck3r):
This function is NOT one-to-one, so there will be no inverse "function"
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OpenStudy (zzr0ck3r):
you can find two different equations to model the inverse image
OpenStudy (jhannybean):
Ohh, because the function does not pass the horizontal line test?
OpenStudy (zzr0ck3r):
correct
OpenStudy (danjs):
ah right,
OpenStudy (zzr0ck3r):
But the original post said nothing about a function, so you can find the inverse relation.
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OpenStudy (zzr0ck3r):
which is what you will get if you keep going like you were before I rudely butted in.
OpenStudy (anonymous):
I got \[y=\pm \sqrt{x-1}\div16\]
OpenStudy (jhannybean):
And the drawing is not to scale by the way, it's just a rough sketch of what it looks like.
OpenStudy (zzr0ck3r):
the 16 should be on the inside \(y=\pm\sqrt{\frac{x-1}{16}}=\pm\frac{\sqrt{x-1}}{4}\)
OpenStudy (anonymous):
ok
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OpenStudy (jhannybean):
|dw:1418797770709:dw|
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