questions

i take the 'ds/'dt, this is my average velocity, then when i need to find my acceleration what do I do here?

On a paper I have, you say to use 'dv not avg vel.

Would I just take my average and double it to get my final and since I know it started from rest take the vf-vo/'t?

When it says find the first 6 terms of the sequence defined by a(n)=a(n-1)+ -3n, a(1)=4. Show that the second difference sequence is constant. So I would just start by saying a(1)=a(0)+-3 a(2)=-2 a(3)=a(2)+ -9 and write here you wrote on my paper saying =-2+(-9)=-11

a(3)=a(2)+ -9 , and from the step before that you know that a(2) = -2, so you get a(3) = -2 + -9 = -11, so a(3) = -11.

I'm not sure what you mean there b/c I thought that you couldn't simplify that anymore than it already was. Anyways and it continues a(4)=a(3)+-12 and here you wrote =-11+(-12)=-23

a(4)=a(3)+ -12 , and from the step before that you know that a(3) = -11, so you get a(4) = -11 + (-12) = -23, so a(4) = -23.

I'm unsure of what you did here also Then a(5)=a(4)+-15 a(6)=a(5)+-18

These two steps follow by the same reasoning as the first few steps.

OK and I understood that some of the pattern was that it increases by a negative 3 but I am still unsure about what the corrections were... I don't know how to get what you got!

Ok when it gives me a quadratic function and asks me to find the vertex and the points 1 unit to the right and 1 unit to the left of the vertex, what exactly is it asking me do to do.

I know that you would just plug the a, b, & c into the -b +/- the square root of b^2-4ac all divided by 2a. Thats how you find the zeros, right? and then to find the vertex it would be -b/2a

Right. The vertex is at x = - b / (2a).

You have to substitute this value into the quadratic to get the y coordinate of the vertex.

So to get the points 1 unit to the right and left of the vertex I would just take whatever I got for the vertex and subtract one from it, and then add one to it. Is this correct?

This is what you would do to get the x coordinates 1 unit right and left.

You also add a to the y coordinate of the vertex to get the y coordinates 1 unit right and left.

This gives you the three basic points of the parabola.

Approximate e using n=1, 1000, 1000000. Knowing that to n decimal places e=2.718281828, how does the number of accurate digits in the approximation change when n increases by a factor of 1000?

To approximate e for a given n you plug that number n into the expression (1 + 1/n) ^ n.

For n = 1000 you get 2.716923932. This differs from 2.718281828 in the fourth significant figure, so is accurate to 3 significant figures.

For n = 1,000,000 you get 2.718281303 This differs from 2.718281828 in the 8th significant figure, so is accurate to 7 significant figures.

For the function y=t^-1.5015 construct a table of y vs. t for t running from t=2.3 to t=2.46 in four equal increments.

Using appropriate transformation(s) on the y column, the t column, or both, linearize this data set and demonstrate that the data set has in fact been linearized. I understand the part about making a table, however, I don't know what it means about approriate transformation(s) and linearizing the data.

See also other notes about linearizing at access page 23-24-242.

To linearize you check to see if log(y) vs. log(x), or y vs. log(x), or log(y) vs. x gives you a straight-line graph. If one does then you find the slope and vertical intercept, write the appropriate equation and solve for y in terms of x.

What are the first five numbers generated by the difference equation a(n+1)=.95a(n), a(0)=4 What exponential function models the function a(n) vs. n Ok I started just plugging in 0, 1, 2, 3, 4, etc. into the equation which gave me, a(0+1)=.95a(0) a(1)=.95a(0) and the a(0)=4 so a(1)=.95(4) - a(1)=3.8 a(1+1)=.95a(1) a(2)=.95a(1).

Right. And since a(3) = 3.8, a(2) = .95 * 3.8.

a(3)=.95a(2)

Yes. Then substitute the value you got for a(2) and calculate a(3).

Then proceed in a similar manner to get the next few values.

a(4)=.95a(3) a(5)=.95a(4)

Construct a graph of y=f(x)=4*1^x and determine the average slope between x=3.5 and x=3.58, and also between x=8.1 and x=8.18.

Determine the approximate average value of f(x) on each interval. Show that the ratio of slope to average value is very nearly the same on both intervals. I don't know exactly what to do with the f(x) in the middle of the equation there, so I'm assuming that you just plug each x value into the equation to get the y value. y=f(x)=4*1^3.5 which would just be y=f(x)=4 and the y would be the same for all these numbers b/c 1 raised to the power of anything is still just one. So we have (3.5, 4), (3.58, 4) and (8.1,4), (8.18, 4). Is this correct? This concludes that there is no rise. Am I on the correct track here?

Everything is right so far. You would conclude that the slopes are all zero.

This is because 1^x is always equal to 1.

If it had been, say, 2^x, or 1.3^x, you would have obtained nonzero rises and nonzero slopes.

Sketch a graph of y=(x+3.5)^2(x-5.5)(5x^2+5x+3.5)

Could you explain this one to me, especially that (x+3.5)^2

Would I think of this of this one as one polynomial or 2 and what would be its zeros? -3.5 or positive 12.25? Do you count the (x+3.5)^2 as a degree 1 or two?

If you multiply out (x+3.5)^2 you get x^2 + 7 x + 12.25. The highest power is 2, so this expression is of degree 2, also called quadratic.

(x + 3.5)^2 is zero if and only if x + 3.5 is zero.

x + 3.5 = 0 if x = -3.5.

(x + 3.5)^2 is just the y = x^2 parabola horizontally shifted -3.5 units, so its vertex is at (-3.5, 0). The rest of the polynomial effectively creates an additional vertical stretch near this point, but the graph of y=(x+3.5)^2(x-5.5)(5x^2+5x+3.5) will be nearly parabolic as it touches the x axis at (-3.5, 0).

(x-5.5) would have a zero value of 5.5 and be a degree 1?

The highest power of x is the first power so the degree is indeed 1. A degree 1 polynomial is a linear function.

The zero does occur at (5.5, 0), and the graph goes straight through this point.

(5x^2+5x+3.5)

Since this has three x's, is it a degree 3? Would I factor this one out? What would be its zero?

The highest power of x is 2 so this is a degree 2 polynomial, also called a quadratic polynomial.

5 x^2 + 5 x + 3.5 = 0 if and only if x = ( - 5 +- sqrt( 5^2 - 4 * 5 * 3.5) ) / (2 * 5).

Since 5^2 - 4 * 5 * 3.5 = -45 is negative, sqrt(5^2 - 4 * 5 * 3.5) is an imaginary number and ( - 5 +- sqrt( 5^2 - 4 * 5 * 3.5) ) / (2 * 5) is a complex, not a real number. So the quadratic factor 5 x^2 + 5 x + 3.5 is irreducible and contributes no zeros to the function y=(x+3.5)^2(x-5.5)(5x^2+5x+3.5).

See also other posted explanations of graphs.

What are the basic points of the exponential function y=f(x)=9*e^x. Graph the function using these points. The basic points are 1,0,-1, correct? I am still confused here when it has the f(x). 9*e^1=24.26 9*e^0=9 9*e^-1=3.31 Is this what you're looking for?

Right. To be very specific:

The points are (-1, 3.31), (0, 9) and (1, 24.26).

Officially (-1, 3.31) isn't a basic point, but it's fine to include it. Instead of a basic point at -1, we understand that as we move to larger and larger negative values of x, e^x approaches zero and gives us a horizontal asymptote with the negative x axis.

So we say that the basic points are (0, 9) and (1, 24.26) with a horizontal asymptote at the negative x axis.

The next question is The population of a country is 73 million in 1902 and 70 million in 3. (I'm assuming you mean 1903) What exponential function models the population in terms of the time t in years since 1980? How long will it take the population to fall to half its original value?

Ok, with this one, I'm assuming a decreasing population, so would I take 70/73 to get the average year decrease?

Good insight. Every year the population will be 70/73 of the previous year's population.

70/73 = .959, approx.. This tells you that the growth factor is .959, and the population function is therefore P(t) = A * .959 ^ t. The value of A has yet to be determined.

If t is the time since 1980, then we know the following:

In 1902 we would have t = -78, since 1902 is 78 years before 1980. In 1903 we would have t = -77, since 1902 is 78 years before 1980.

So if population is understood to be in millions, we have

73 = A * b^-78 and 70 = A * b^-77.

If we divide the second equation by the first we get

70 / 73 = A * b^-77 / (A * b^-78) so that 70 / 73 = b^(-77 - (-78) ) or 70/73 = b^1.

This tells us that b = 70 / 73 = .959, which we already knew.

Substituting this into the first equation we have

73 = A * .959 ^ -78 so that A = 73 / (.959^-78) = 26.4, approx..

Thus our function is

P(t) = 26.4 * .959^t.

Sketch a graph of the polynominal -1(x-2)(x+3)(x-8), clearly indicating all intercepts and the large -[x] behavior of the graph. Ok I am unsure of what exactly the negative 1 in front of the polynomials does. I know it crosses the x axis at 2, 8 and -3. This is a degree three polynomial?

The degree of the polynomial is the highest power of x in its expanded form. If you expand this polynomial the highest power of x will be 3, so the polynomial is of degree 3.

If x is a large negative number, then x -2, x+3 and x - 8 are also large negative numbers and -1(x-2)(x+3)(x-8) is a product of -1 and three large negative numbers. The product of four negative numbers is positive and the product of three large numbers is huge, so the value of the polynomial will be a huge positive number.

If x is a large positive number, then x -2, x+3 and x - 8 are also large positive numbers and -1(x-2)(x+3)(x-8) is a product of -1 and three large positive numbers. The product of one negative number and three positive numbers is negative and the product of three large numbers is huge, so the value of the polynomial will be a huge negative number.

Thus, starting from the left, the zeros are at -3, 2 and 8. So the function -1(x-2)(x+3)(x-8) descends from very large positive values to the first zero at (-3, 0), then turns around and comes back up through its second zero at (2, 0), then finally turns around and comes back down through its third zero at (8, 0) before descending rapidly through negative values as it continues to the right.