- Sun Mar 17, 2013 1:06 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solving exponential equation y = e^(-x) -x + 1
- Replies:
**3** - Views:
**4893**

Well this is a calculator paper so I suppose I can use a graph...but the answer says 'Attempt to solve this numerically' so I thought there would be a way to solve this.

- Sun Mar 17, 2013 2:18 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solving exponential equation y = e^(-x) -x + 1
- Replies:
**3** - Views:
**4893**

y = e^(-x) -x + 1 I am told to find the x-intercept so: 0 = e^(-x) - x +1 I have no idea what to do next, I tried to cancel out the denominator by multiplying everything by e^x so it becomes 1 - xe^(x) + e^(x) = 0 1 - e^(x) [x+1] = 0 I am stuck here, can anyone please help me out a bit here :confused:

- Mon Jan 14, 2013 11:21 am
- Forum: Trigonometry
- Topic: Graphing a trig function
- Replies:
**4** - Views:
**4873**

So you mean I just plug in pi/6, pi/4, pi/3 etc...? Because I was kind of looking for a solution instead of using the substitution method

- Sat Jan 12, 2013 12:49 pm
- Forum: Trigonometry
- Topic: Graphing a trig function
- Replies:
**4** - Views:
**4873**

The question is as follow: f(x) = (2+3sinx)/(4+3cosx) I know that the first thing you do is to try simplifying this down to 1 trig ratio, but well, I use cos(x) = sqrt[1-sin^2(x)], which got me nowhere. I have looked at the answer and the shape appears pretty weird to me so I have no idea where to s...

- Wed Jan 02, 2013 5:25 am
- Forum: Trigonometry
- Topic: Finding the phase shift
- Replies:
**1** - Views:
**5178**

I understand that in asin(bx + c) + d, where c represents the phase shift. But then in a question like: f(x) = 7sin[6(x + pi/12)] + 3 the phase shift here is pi/12 (I thought it was pi/2) and f(x) = -3sin(2x + pi/2) - 5 the phase shift here is pi/4...again I thought it was pi/2. So how exactly do yo...

- Mon Nov 26, 2012 5:04 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithm and infinite geometric series
- Replies:
**5** - Views:
**5297**

I realized I could use the ratio test to solve that sequence, although I have yet to learn it at school. But then I am still confused with the a + log (base5) b part In fact this is what I got: 5^x = 3/5 x = log(base5)^3/5 = log(base5)^3 - log(base5)^5 = -1 + log(base5)^3 so a = -1 and b = 3? I don'...

- Sun Nov 25, 2012 12:31 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithm and infinite geometric series
- Replies:
**5** - Views:
**5297**

No, that is the entire question.

- Sat Nov 24, 2012 2:03 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Logarithm and infinite geometric series
- Replies:
**5** - Views:
**5297**

"Consider the infinite geometric series 1 + (2x/3) + (2x/3)^2 + (2x/3) ^3 +... For what values of x does the series converge?" I know convergence means the sum tends to a finite series, so I have used the equation Sn = u/(1-r) where u = first term and r = ratio Sn = 1/(1-2x/3) But then I have no ide...

- Sun Sep 16, 2012 10:17 am
- Forum: Intermediate Algebra
- Topic: Polynomial Functions Questions
- Replies:
**3** - Views:
**7674**

So r(-1) = -4. Since we are dividing by a quadratic, then logically the remainder can have a degree of 0 or 1. Since r(x) is clearly not constant, then r(x) must be linear. Then r(x) = ax + b for some values a and b. One may then find the equation of the remainder as being r(x) = (3/2)x - (5/2). I k...

- Thu Sep 13, 2012 11:39 am
- Forum: Intermediate Algebra
- Topic: Polynomial Functions Questions
- Replies:
**3** - Views:
**7674**

Hello, Thanks for taking your time reading. I have encountered 2 problems as I was going through my textbook, so may I please kindly ask of some assistance on this. "Find the polynomial f such that f(2x-1) = 16x^4 - 32x^3 +12x^2" And "When polynomial f is divided by (x-3) the remainder is 2, and whe...