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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) = \frac{1}{2} x^2 - 4 $

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In Exercises 7-12, match the quadratic function with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] $ f(x) = (x + 1)^2 - 2 $

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In Exercises 7-12, match the quadratic function with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] $ f(x) = (x - 2)^2 $

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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) = (x - 1)^2 $ (b) $ g(x) = (3x)^2 + 1 $ (c) $ h(x) = \left(\frac{1}{3} x^2 \right) - 3 $ (d) $ k(x) = (x + 3)^2 $

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Okay, so this is going to be the fourth example out of our determinants, Siri's and it says, find the determinant. So currently we have a three by three matrix, and we know that because we have three rows as well as three columns. So I know that this one I'm good. I can set it up as a certain a times a two by two matrix minus B times a different two by two matrix plus C times a different two by two matrix. And that is extrapolated from if our three by three matrix is a B c d e f g h i. So the first row is like a key value. We have to use that to find the rest of our values. So I already know that in my a B and C positions Aiken substitute in the values negative to positive one. And negative, too. I'm gonna go on and delete that. Plus, So then, from here, the way to do it is I'm going to circle that first value negative two, and then I'm going to cross out, um, the associate ID row and column. So then what's remaining is this component right? here, Which is this? A two by two matrix. So the two by two matrix that's formed is five negative five to negative five. So then I'm going to repeat that same process for the second one. Um and I'm gonna go ahead and circle that one and anything that is in that same row, which is this in the same column, which is this I'm gonna go ahead and get rid of, and then the resulting is gonna become its own two by two matrix. So zero negative 50 and negative five. So then again, same type of process. We're going to go ahead and circle that negative two and anything that's in that same row or column I'm going to cross out. And so the resulting matrix is going to become 0502 So now I have the template that we want. And then from here I have Thio basically within finding the determined of a three by three matrix, we have to know how to find the determined of a two by two matrix. So be sure that you understand how to do that. And that is described within not only the lecture video for determinants, but also in the first lecture video of determinants or the example video of determinants. So I'll show you how to do it two by two. But in general, with a two by two, we know that we multiply this section here and then subtract that. Subtract this section that's multiplied. So if we were to think about a two by two matrix being a B C D, it would be a D minus b. C. So you multiply and d on the subtract B times C. So if we do that to the first set, I get negative two times negative five times positive, five minus negative, two times negative five. And then I have to subtract the negative one times, zero times negative. Five, um, minus zero times Negative five. And then I'm gonna add the negative two times the zero times too minus the zero times five. So then if I try to simplify that, I would get negative two times negative. Five times five is negative. 25 and then negative. Two times negative five becomes positive. 10 since two negatives multiplied to become one positive. I got minus 10 and then this component right here that become zero because zero times anything become zero, and then zero times anything also become zero. I'm going to say minus zero and then this component also zero times anything become 00 times anything that become zero. So if I have zero for this entire thing zero times negative to also become zero, so it's minus zero again. So there were only really concerned with this first component. So for that one, we know that we've got negative two times. Negative. Let's see, What does it become? Negative two times? Mm hmm. Negative. 35 It looks like So that's going to become positive. 70 actually, like this one was actually a positive to from here so that if we bring that down, that becomes ah plus right here because it was negative. Two times negative. Five becomes negative. 10 here. So negative, negative becomes positive. So we've got that when that becomes negative. 15. So this actually becomes positive. 30. So our determinant of this three by three matrix is positive. 30. So then just to go back and recap all the steps that we did, we have to fill in that certain format of eight times a two by two minus B times 22 by two plus C by that two by two. So keep in mind that this the signs are important. So the minus comes first and then the plus. And then once we have that format, we have to take the determinants of all of them and then multiply that by the A, B and C values. And then finally, we just have to simplify, right?

Introduction to Conic Sections

Discrete Maths

Introduction to Combinatorics and Probability

Introduction to Sequences and Series

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