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Yasemin E.
July 17, 2021
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?Mcg? ?.
October 18, 2021
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Campbell University
University of Michigan - Ann Arbor
Boston College
Utica College
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J H.
In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s). $ f(x) = x^2 + 7 $
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In Exercises 13-16, graph each function. Compare the graph of each function with the graph of $ y = x^2 $. (a) $ f(x) = \frac{1}{2} x^2 $ (b) $ g(x) = -\frac{1}{8} x^2 $ (c) $ h(x) = \frac{3}{2} x^2 $ (d) $ k(x) = -3x^2 $
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In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s). $ h(x) = 12 - x^2 $
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In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s). $ g(x) = x^2 - 8 $
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Okay. Okay. So this is going to be the first example out of our continuity. Siri's and the question says, determine if each function is continuous and if not continuous, classified each discontinuity. So first of all, let's think about Okay, Well, what does it even mean for function to be continuous for something to be continuous, that means that we condone draw like, basically trace that function without ever taking off my pencil. So, for example, if I had a function, it looked like something like, let's just say something like this. So just speaking on continuity, if I have something like that, well, I could technically take my pencil and trace it without ever lifting up my pencil. So, technically, this is continuous. It wouldn't really be a function in that it doesn't pass a vertical line test. But just speaking on continuity, um, that would work. So then what would not be considered OK is if we have something like, um, maybe something where we have this. And then there's a singular asked him to appear. Maybe something like this. Well, that doesn't work, because we get to hear and we have toe lift up our pencil and then jumped to here and then start drawing again because we can't actually put anything within that circle because that's what that circle means. It's a singular aspecto, but that's what would mean. That's what it would mean for something to be discontinuous. So let's just look at the equation that we have. Well, we know that we have. Ffx is equal to X over X squared plus three X, and so basically, you're thinking about what kind of accent hopes might be there, whether that be vertical ascent, Taub's Horizontal, Allison Taub's Slammed Ascend, Taub's or singular Aston Taub's Well, in this situation, we know that we have, um, we cannot have a zero as a denominator, no zeros in denominator. That doesn't mean that the X value camp easier necessarily. It just means that the entire denominator cannot equal zero. So if the entire denominator cannot equal zero, let's think about when it does equal zero and then exclude those points. So it's kind of like saying Okay, well, what are you allergic to? Let's take those out and then make the cake. So then, if we say that we have X squared plus three X If we set that equal to zero. You know, we can factor out the X from both terms leaving us with X plus three is equal to zero. So this right here is my factored form, Then using my factor form. I know that, Teoh, in order for either one of these to become zero, either the factors could become zero, at which point excess, either zero or negative three. So that must mean that if I plug in either of those values, the entire denominator is going to become zero, which, like I said, is exactly what we want to avoid. So since we're trying to avoid that, we know that there's going to be a vertical ascent. Ope at X is equal to zero, and X is equal to negative three eso. Actually, it's gonna be a X is equal to negative three. Because now that I think about it, if we have X over X Times X plus three, the top X and the Bottom X would technically cancel out. So we would only have a singular point here. So I'm going to erase this, and I would say that I have a point. Assad took singular quite a Scinto X equals zero. So we've got to absent hopes. Definitely one singular and one vertical on bond. So because we have any aspect hopes at all, we know that we can't draw that function without taking off my pencil. So the function is discontinuous, and it is discontinuous at these two particular points.
Powers and Polynomial
Rational Numbers
Logarithms
Exponential Functions
Trigonometry
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