WPP #6: Unit I Concept 3-5- Compound Interest

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SV#4: Unit I Concept2: Graphing Logarithmic Functions

One thing your going to want to look out for in my equation and probably most equation like this, is that when you plug it in the equation, use the change of base formula. It will make things easier and get the most accuate graph. If you change to the table you can also find key points.

WPP #2: Unit A Concept 7 - Profit, Revenue, Cost

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WPP #1: Unit A Concept 6 - Linear Models

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SP#3: Unit I Concept 1: Graphing Exponential Functions

Equation: You can see that -4 is A, 2 is B, 3 is H and it comes from (x-3) which is the exponent of 2, and -1 for K.
Key Points: I've used the -2, 1, 2, and 3 for the x-intercepts. To get the y-intercept you need to plug in equation I've given you into your calculator. But you'll need to use the change of base formula to get the right points.
Asymptote: To get the asymptote is pretty simple, it's just the number in K's place. In mine it will be -1.
X-intercept: To find X you will need to replace Y with 0 and then add 1 to both sides. Next divide -4 to both side and you should have -1/4 = 2^(x-3). Last you'll put LN to both sides to get ride of the 2. But in this case there is no x-intercept because LN can not be a negative number.
Y-intercept: First you'll have to replace X with 0 and the exponent for 2 will be -3. 2 with it's new exponent will now be 1/8. If you multiply the -4 to 1/8 you'll get -1/2. Add -1 to -1/2 and your answer should be -3/2. And your y-intercept should be (0,-3/20.
Domain and Range: The domain will be infinite for the x and y because it goes left and right. The range will be negative infinity for the x and -1 in the y because of the asymptote on -1.

In my problem the parts you need to pay special attention to are the x and y-intercepts. X because of the -4 and the fact that we can't have negative LOGs. Also Y due to the fact that there are fractions and it'll get with that.

SV#3: Unit H Concept 7: Finding Logs Given Approximations

     Well one thing you should look out for in this problem is knowing that there is a free clue. Some problem problems might not even use them but mine does. Also when it's already in exponential form, the LOGs that have exponents, their exponents should be moved to the front of the LOG so it can be the leading coefficient.

SV#2: Unit G Concepts 1-7 - Finding all parts and graphing a rational function

     This problem is about finding three different asymptotes; horizontal, slant, and vertical.To have a horizontal there can't be a slant, but in my problem there is a slant but no horizontal. It does have a vertical when we foil out the numerator and the denominator and (x+7) cancels because it's found in both. So this problem also has a holes. But once we have all of those we find the domain, x-intercepts, and y-intercept. Then whats left is the graphing.
     In my problem one thing you'll have to pay close attention to the hole points. They can be tricky because you may never know where they go. On my it's pretty hard to tell where the hole goes.