![]() ![]() Negative numbers, so 3 to the negative thousand, 3 to the Larger negative numbers, or I guess I should say smaller It's going to get closerĪnd closer to zero. If I did 5, we'd go toĢ43, which wouldn't even fit on my screen. When x equal to 3, y isĮqual to 27, which is right around there. So when x is equal to 0, we'reĮqual to 1, right? When x is equal to 0, y is equal So let me draw it down hereīecause all of these values, you might notice, are positive Let me draw it a littleīit differently than I've drawn it. So let me just draw as straightĪ line as I can. Increments of 5, because I really want to get the general To get an idea of how quickly we're exploding. Power, right? y is equal to 3 to the second power. Larger, a little bit larger, but you'll see that we Have 3 to the 0 power, which is just equal to 1. And then 3 to the negativeĢ power is going to be 1/9, right? 1 over 3 squared, and then we Super-small number to a less super-small number. To the third power, which is equal to 1/27. When x is equal to negativeĭifferent color. The negative 4 power, which is equal to 1 over 3 to Y-values are going to be for each of these x-values. See how quickly this thing grows, and maybe we'llĮqual to negative 4. ![]() The third power, this is 3 to the x power. So let's just write an exampleĮxponential function here. Really just show you how fast these things can grow. Introduce you to the idea of an exponential function and This shifts from the origin to (-2,-5) which makes the asymptote at y=-5, but it is a little harder to determine the x axis shift back 2. This is a good introduction, which is good for all but shifts in the x direction such as y = 3 (4)^(x+2) - 5. Thus, you would have to do (5- 3)/(4 - 3) to get 2/1=2 as the base. Thus y=2^x + 3 would have points (0,4) 1 away from asymptote, (1,5) two away from asymptote, etc. If you see an asymptote at say y=3, then "act like" this is the y axis and see how far points are away from the this line. This can be easily be determined by a change in the asymptote. So the next easiest is to shift up and down by adding a constant to the end. This will work the same for decay functions, but the base will be a fraction less than 1. Similarly, if we have (0,3) and (1,6) our base is 6/3=2, but the scale factor is 3, so we have y=3(2)^x. With 2(2)^x, you double all the y values to (0,2)(1,4)(2,8)(3,16) - note that 16/8=8/4=4/2=2, so we still get the same base, but the y intercept tells us that the scale factor is 2. Next, if we have to deal with a scale factor a, the y intercept will tell us that. It works the same for decay with points (-3,8). Thus, we find the base b by dividing the y value of any point by the y value of the point that is 1 less in the x direction which shows an exponential growth. So we find the common ratio by dividing adjacent terms 8/4=4/2=2/1=2. You start with no shifts in x or y, so the parent funtion y=2^x has a asymptote at y=0, it goes through the points (0,1) (1,2)(2,4)(3,8). Learning the behavior of the parent functions help determine the how to read the graphs of related functions. So one basic parent function is y=2^x (a=1 and b=2). Increasing above that.So the standard form for a quadratic is y=a(b)^x. Over here, I'm not actually onĠ, although the way I drew it, it might look like that. The whole curve, just to make sure you see it. To a super huge number because this thing is just To the positive billionth power, you're going to get To get you to 0, but it's going to get you So you could keep goingįorever to the left, and you'd get closer andĬloser and closer to 0 without quite getting to 0. Keep this curve going, you see it's just going Some people would call itĪn exponential increase, which is obviously the The exponential is good at, which is just this Increasing beyond 0, then we start seeing what Slightly further, further, further from 0. Negative direction we go, 5 to ever-increasing Happens with this function, with this graph. So that right over thereīe right about there. And then my y's go all the wayįrom 1/25 all the way to 25. ![]() My x's go as low as negativeĢ, as high as positive 2. Then y is 5 squared,ĥ to the second power, which is just equal to 25. Let me extend this tableĪ little bit further. So this is goingĬouple of more points here. To the 0-th power is going to be equal to 1. To 5 to the 0-th power, which we know anything Negative 1 power, which is the same thing as 1 over 5 To the positive 2 power, which is just 1/25. Then y is equal toĢ power, which we know is the same thing as 1 over 5 Reasonably negative but not too negative. So let's try some negativeĪnd some positive values. Some values for x and see what we get for y. ![]()
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