OK, I'm having some serious problems with my email, I can't find my access code anymore b/c they are doing something weird to the system, I can't even log into my email through vhcc's page,I finally found another link at southwest's web page and managed to get in there only to find that all my email w/ all the important links and access codes you'd given me were deleted. It gave a system's noticed that we'd have till the 30th to foward them, but looks like they got an early start on mine. I was wondering whether or not you could resend me the directions and access codes THANKS!
I just sent you copies of your codes.
I have been working on a pratice precal test, I worked out some problems but I have a few questions, so here goes.
Ok when I put data into quadratic formula and I get a negative under the square root, would I just put that no such x exists and be done with it or what?
If there is a negative under the square root, there is a complex-valued solution of the form a + b i, where i = sqrt(-1) is the basic 'imaginary' number.
However there are no real-valued solutions.
You can get away with saying that there are no solutions, but its better to say that there are no real solutions. Meaning, basically, that it wouldn't happen in the real world.
The second problem states evaluate f(17.07933) and f(2a+b) for the function y=f(t)=.03t^2+-2.1t+47
What is the value of the function for clock time t=25.619? What equation would you solve to determine the value of t for which f(t)=17.7782?(You need not actually evaluate the equation).
ok so heres what I did,
y=.3(17.07933)^2+ -2.1(17.07933)+47
This is correct.
By contrast note that the solution to the equation f(t) = 17.8 is obtained by solving the equation
17.8 = .03t^2+-2.1t+47.
You solve this by subtracting 17.8 from both sides then using the quadratic formula.
y=.3(2a+b)^2+ -2.1(2a+b)+47
Do I need to actually go ahead and completely solve these out? Or are they ok left as is to start w/?
You can probably get away with the second expression y=.3(2a+b)^2+ -2.1(2a+b)+47 but you should ideally also be able to expand it. The result would be
y=.3(2a+b)^2+ -2.1(2a+b)+47 =
.3 ( 4 a^2 + 4 a b + b^2) - 2.1 ( 2a + b) + 47 =
1.2 a^2 + 1.2 ab + b^2 - 4.2 a - 4.2 b + 47.
When is asks what is the value of the function for clock time t=25.619? I put y=.3(25.619)^2+ -2.1(25.619) +47 and that equals y=190.1
This is correct.
The fourth question states:
If water depths of 37.6, 31, 27.1,and 25.9cm are observed at clock times 17.8, 26.7, 35.6, and 44.5sec, then at what average rate does the depth change during each time interval?
Is the average rate of change during each time interval just slope? How would I set these up just like ... 37.6/17.8 Would that be the average rate of change? Would I just do that for each one?
The average rate of change of depth with respect to clock time is (change in depth) / (change in clock time).
There are 3 intervals, one from clock time 17.8 sec to 26.7 sec, on from 26.7 sec to 35.6 sec, and one from 35.6 sec to 44.5sec.
The change in clock time for the first interval is 26.7 sec - 17.8 sec = 8.9 sec, and the change in depth is 31 cm - 37.6 cm = - 6.6 cm. So the average rate of depth change in this interval is
ave rate = -6.6 cm / (8.9 sec) = -.73 cm / sec, approx. (check my arithmetic).
On a graph of depth vs. clock time the first two points would be (17.8 sec, 37.6 cm) and (26.7 sec, 31 cm). The rise from the first point to the second would be -6.6 sec and the run would be 8.9 sec. So the average slope between these points would be -6.6 cm / (8.9 sec) = -.73 cm/s.
Since the rise represents the change in depth and the run represents change in clock time, the slope represents the average rate of change of depth with respect to clock time.
You can do the same thing for each of the remaining two intervals.
On the second the average rate would be -3.9 cm / (8.9 sec) = -.45 cm/sec, approx., and this would be equal to the slope between the second and third graph points.
On the third the average rate would be -1.2 cm / (8.9 sec) = -.14 cm/sec, approx., and this would be equal to the slope between the second and third graph points.
I know that if you gave me a list of the time and depth changes I would just get the difference in time changes and take the depth/time changes... But if you don't give me the depth changes would I just do the same thing?
You wouldn't want to take the difference between the depth changes, but the difference between the depths. Same with times--difference between clock times, not difference between changes in clock times.
Then it states sketch a graph of this data set and use a sketch to explain why the slope of this graph between 26.7 and 35.6 represents the average rate at which depth changes during this time interval?
This is explained by explaining what the rise and the run mean, as above.
and finally if f(x)=x^2, give the vertex and the three basic points of the graphs of f(x-1.75), f(x)-.85,2 f(x) and 2f(x-1.75)+.85
I know how to find the vertex if I have a model, but without it I'm unsure of what to do!
The vertex of the basic function f(x) = x^2 is (0, 0). I know that you know how to sketch the graph of the basic y = x^2 function and see that the vertex is at this point.
The basic points are the x = -1, x = 0 and x = 1 points (-1, 1), (0, 0) and (1, 1).
f(x - 1.75) replaces x with x - 1.75, so that to get the same y values as the basic function your x values have to be 1.75 units greater. This shifts the graph 1.75 units to the right, so that the basic points shift as follows:
(-1, 1), having x value -1, shifts to (-1 + 1.75, 1) = (.75, 1)
(0, 0), having x value 0, shifts to (0 + 1.75, 1) = (1.75, 0)
( 1, 1), having x value 1, shifts to (1 + 1.75, 1) = (2.75, 1).
These are the basic points of the graph of f(x - 1.75). Plot them and you can easily sketch the parabola defined by these points.
y = f(x)-.85 has y values .85 less than y = f(x). That is, the graph is shifted downward .85 units. The basic points shift as follows:
(-1, 1), having y value 1, shifts to (-1, 1 - .85) = (-1, .15)
(0, 0), having y value 0, shifts to (0 , 1-.85) = (0, -.85)
( 1, 1), having y value 1, shifts to (1, 1 - .85) = (1, .15).
These are the basic points of the graph of f(x) - .85. Plot them and you can easily sketch the parabola defined by these points.
y = 2 f(x) has y values that double those of y = f(x), stretching the graph by a factor of 2 in the y direction. The basic points shift as follows:
(-1, 1), having y value 1, shifts to (-1, 2 * 1 ) = (-1, 2)
(0, 0), having y value 0, shifts to (0 , 2 * 0) = (0, 0)
( 1, 1), having y value 1, shifts to (1, 2 * 1 ) = (1, 2).
These are the basic points of the graph of 2 f(x). Plot them and you can easily sketch the parabola defined by these points.
The function 2 f(x - 1.75) + .85 first shifts the function 1.75 units in the x direction, then stretches the graph by factor 2 in the y direction, and finally shifts the graph .85 units in the y direction. In the process
(-1, 1) shifts in the x direction to (.75, 1), then stretches in the y direction to (.75, 2) and finally shifts in the y direction to (.75, 2.85).
(0, 0) shifts in the x direction to (1.75, 0), then stretches in the y direction to (1.75, 0) and finally shifts in the y direction to (.75, .85).
(1, 1) shifts in the x direction to (2.75, 1), then stretches in the y direction to (2.75, 2) and finally shifts in the y direction to (.75, 2.85).
These are the basic points of the graph of 2 f(x - 1.75) + .85. Plot them and you can easily sketch the parabola defined by these points.
And the last question I have is about
Find t such that f(t)=6.903577 and find f(t) when t=33, given the depth vs. clock time function
y=f(t)=.021t^2+ -2.61 t +87.
Using the same function determine the clock time when water depth is 43.90358, the depth at clock time t=12
I can do that last part, you just substitute the time in for t to find the depth, and to find the clock time you just set the equation equal to the depth and then get it equal to zero and then substitute what you get into the quadtratic equation.
Right. See also my preceding notes.
I'm unsure of that it wants in the first part. Thanks for all the help, I think after I get these problems figured out that I'll be ready for the test! Thanks!
To find t such that f(t)=6.903577, since y=f(t)=.021t^2+ -2.61 +87 you get the equation
6.90 =.021t^2+ -2.61 t +87, which you solve as a quadratic equation.
To find find f(t) when t=33 you plug t = 33 into the function y=f(t)=.021t^2+ -2.61 +87 to obtain
y=f(t)=.021* 33^2+ -2.61 * 33 t +87 and just evaluate the expression.
Also since I can't view my questions and answers can you email them back to me as well so I can review them! Thanks
sure will
See also the related question below:
If you're is supposed to graph y=2^x and then stetch it vertically by factor 1.29 units and then shift it -2.61 units, how would you write that as an equation? y=(2-1.29)^x-2.61, is that correct?
To stretch vertically by factor 1.29 you multiply y values by 1.29. This would change the basic function
y = 2^x
to the function
y = 1.29 * 2^x.
Assuming the shift is horizontal, i.e., in the x direction, you would then replace x by x - h with h = -2.61. Thus x would be replaced by x - ( - 2.61) = x + 2.61 and your function would become
y = 1.29 * 2^(x + 2.61).