(Solved) Sketch the graph of each function using transformations. State the domain and range.

marinejas

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3 years ago Edited: 3 years ago, marinejas

Sketch the graph of each function using transformations. State the domain and range.

a) f(x) = 3 log10(bottom exponent) x + 3
e) k(x) = 4 log10(bottom exponent) (1/6x) - 2
f) r(x) = log10(bottom exponent) (-2x -4)
Hi so I had some trouble creating a chart with these 3 different equations, the other video that you suggested showed mainly the basic ones like y = 5^x, so just wanted to know if you know how to graph these ones, y=a log 10(bottom exponent) [k(x - d)] + c thanks, if not that's okay as well.

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bio_man

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bio_man

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Educator

3 years ago

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marinejas Author

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3 years ago

okay that makes sense, and say if I wanted a more accurate graph with more points, I would not only use the original points (1,0) and (10,1) but also (1/10, -1) and (32, 1.5). If I wanted to use those last two points, I would just do the same exact same?
Overall when there's a vertical stretch you would always multiply (a value) and when its the c) value you subtract/or add (whether it goes up or down). I'm just a little confused on how you did the horizontal stretch.

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bio_man

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3 years ago

For the horizontal stretch, I don't know any other way but to plug the x-value into the equation provided.

You're right about all those other follow-up questions. You're free to use as many points as you like.

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marinejas Author

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3 years ago

The x value of the points (1, -2) (10,2) into the equation? Because I wasn't getting (1, -5.1) (10, -1.1).

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bio_man

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3 years ago

For the second equation, you should if you sub. x = 1 then 10.

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marinejas Author

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3 years ago

okay got it now, but just another thing, so the x values of these points never change? They just affect the y's?

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bio_man

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3 years ago

Yes, x is independent, y is the dependent.

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marinejas Author

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3 years ago

Okay thanks again bio_man, I'm really sorry but just another question before I mark this solved. How about if I was given a horizontal translation? What would you do.

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bio_man

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#10

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3 years ago Edited: 3 years ago, bio_man

log (x + a)

The "a" part?

So if we start with (1,0), and a is + 6, then the new point would be (-6, 0). However, if you plug in x = 1 into the equation, you'll get a new y-coordinate as well for 1.

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bio_man

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#11

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3 years ago

I meant to write (-6,0)

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marinejas Author

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#12

3 years ago

I tried doing f) and I started by -2x which is a horizontal compression by a factor of -1/2. So I plugged in 0.1, the x value (my first point) into the equation log(-2(0.1)-4) but gave me an error on my calculator.

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bio_man

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#13

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3 years ago

That's expected marinejas:

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