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Show that the curve $ y = \sqrt{x^2 + 4x} $ has two slant asymptotes: $ y = x +2 $ and $ y = -x - x $. Use this fact to help sketch the curve.

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Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

Derivatives

Differentiation

Volume

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Use the guidelines of this…

So for this problem, we first want to consider what the graph is going to look like. So when we take the square root of an X squared function, it gives us something like the absolute value functions we see right here. And then we add before X it will be very similar. The only difference is that it's going to separate the range at which we can have values. It's going to change the domain slightly right here. We saw it, shifted it for units to the left, and it caused this big gap right here. So all these values are undefined but were mainly focused on the slant assim totes. We noticed that there are two slant ascend totes as the problem suggests, um, and they're going to go along this line right here in this line right here. So one of them is gonna have positive Slope One is gonna have negative slope. If we look at why equals X, for example, we see that the slope appears to be good. Um, but we're definitely gonna want thio shifted upwards. So well, dio X plus one, we're standing shifted. Absolutely. Experts to this appears to be an awesome toe. And as we go out closer to infinity, we see that they never intersect. There's never a point of intersection. Okay, then we can look the other way. It's going to be somewhat symmetrical. So it's gonna be why equals negative X? Um, my equals negative X. And once again, we're going to need Thio Get it closer that line because there's still a bit of distance. So when we do minus two, we see that it's approaching that line at a much closer distance. However, as we go out to infinity, we see that it still is never intersecting. Therefore, we do have to slay Assam totes X Plus two and negative X minus two.

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