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Problem 53 Hard Difficulty

Find an expression for the function whose graph is the given curve.

The bottom half of the parabola $ x + (y - 1)^2 = 0 $

Answer

We need to solve the given equation for $y . \quad x+(y-1)^{2}=0 \Leftrightarrow(y-1)^{2}=-x \Leftrightarrow y-1=\pm \sqrt{-x} \Leftrightarrow$ $y=1 \pm \sqrt{-x} .$ The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want $f(x)=1-\sqrt{-x}$. Note that the domain is $x \leq 0$.

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Video Transcript

Okay, so we want to find a function that represents the bottom half of this parabola. So when you see an equation like this, it's going to be a problem that opens to the right or opens to the left. Let's just say it opens to the left. And so what's gonna happen is you're going to get if you take the equation and isolate why you're going to get something with a plus minus. And the minus represents the bottom part, and the plus represents the top part. And so let's go ahead and isolate why in this equation and see what we get. So we'll start by subtracting X from both sides, and we get why minus one quantity squared equals the opposite of X. And then we can square root both sides, and we get why minus one equals plus or minus the square root of the opposite of X. And then we can add one to both sides, and we get Y equals one plus or minus the square root of the opposite of X. And again, the one with a plus. That's going to be the top half of the parabola, the one with a minus. That's going to be the bottom half of the parabola. So the bottom half is why equals one minus the square root of the opposite of X.