1008

Do the Introduction to Proportionality worksheet on your Assignments Page. Actually do the experiment as given there.

You may email me for feedback on your test. I expect to be able to answer emails this weekend but cannot guarantee a response before Monday.

Be sure to see the 10/11 page for instructions for Monday.

If f(x) = 2 x - 5 then what are the symbolic expressions for the following (simplify if possible):

( f(x2) - f(x1) ) / (x2 - x1)

f(x2) = 2 x2 - 5 and f(x1) = 2 x1 - 5 so the expression is

( 2 x2 - 5 - ( 2 x1 - 5) ) / (x2 - x1) = ... = 2.

The details of the simplification:

( 2 x2 - 5 - ( 2 x1 - 5) ) / (x2 - x1), to which we apply the Distributive Law to get
(2 x2 - 5 - 2 x1 + 5 ) / (x2 - x1). Since -5 + 5 = 0 we get
(2 x2 - 2 x1) / (x2 - x1). Avoiding the temptation to 'cancel' incorrectly we factor 2 out of the numerator to get
2 ( x2 - x1) / (x2 - x1). Since (x2 - x1) / (x2 - x1) = 1 we have
2 * 1 which gives us our final result
2.

This expression represents the slope of the line segment between the points (x1, f(x1) ) and (x2, f(x2) ), since `dy = f(x2) - f(x1) and `dx = x2 - x1 so that `dy / `dx = ( f(x2) - f(x1) ) / (x2 - x1).

Since f(x) in this case is a linear function the equation of the function gives us the slope, which for f(x) = 2 x - 5 is clearly 2. So we better get 2 when we simplify this expression.

f(x2 - x1)

f(x) = 2x - 5 so

f(x2 - x1) = 2(x2 - x1) - 5 = 2 x2 - 2 x1 - 5.

This is not the same as f(x2) - f(x1).

f(x2) / f(x1)

f(x2) / f(x10 = ( 2 x2 - 5) / (2 x1 - 5).

Nothin' cancels, nothin' simplifies.

(f(x + `dx) - f(x) ) / `dx

f(x + `dx) = 2 ( x + `dx) - 5 so

f(x + `dx) - f(x) = 2 ( x + `dx) - 5 - (2x - 5) = ... = 2 `dx so

( f(x + `dx) - f(x) ) / `dx = 2 `dx / `dx = 2.

Quiz Results from Wednesday's class with some notes.

Approximately 70% of the quiz questions were answered correctly.

The table below shows questions for which the wrong answer was given by students on today's quiz. If an answer was given by more than one student then it will appear more than once.

You should be sure you understand the error that led to each incorrect answer.

Correct answer Question

n The zeros of the function y = f(t) are found by plugging 0 in for t and evaluating the function to find y.
*&*& You find the zeros by setting f(t) equal to 0 and solving for t. *&*&

y There are parabolic functions for which the vertex is the lowest point.

y If the graph of a quadratic function passes through the horizontal axis it does so at horizontal coordinates (-b +- sqrt(b^2 - 4 a c) ) / (2a).

y For a function y = f(t), there can be only one value of y which corresponds to t = 0.
*&*& Plug in t = 0 you get y = f(0), whatever that equals. That's all you get. We're implicitly assuming that 0 is in the domain of the function, and that should be stated in the problem statement.

Don't confuse with the fact that a function can have many zeros. That's where y = 0, not where t = 0. *&*&

n For a function y = f(t), there can be any number of values of y which correspond to t = 0.

*&*& Not so. There is exactly one value, assuming that 0 is in the domain of the function. *&*&

y The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis. If t is a valid value of the independent variable (i.e., if t is in the domain of the function) then there will be exactly one such point.

n The values of t for which the function y = f(t) is equal to c are found by evaluating the expression f(c).
*&*& y = f(t) is equal to c if y = c, not if t = c. *&*&

n The equation f(t) = c might have any finite number of solutions but cannot have a infinite number.
*&*& for example a sine-wave type function can take any y value in its range and infinite number of times. *&*&

n The values of y which correspond to t = 0 are represented by the points where the graph of the function crosses the t axis.
*&*& t = 0 points are on the y axis, not necessarily the t axis. A t = 0 point is your y=intercept. *&*&

y The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis.

y The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis. If t is a valid value of the independent variable (i.e., if t is in the domain of the function) then there will be exactly one such point.

n If a data set such as [ [2, 9], [3, 12], [4, 18] ] is in line #20, then the command fit( [x, a x^2 + b x + c ] , #20) can be simplified to give you the equation of the parabola that best fits the data.
*&*& fit( [x, a x^2 + b x + c ] , #20) tells Derive that the form is a x^2 = b x + c and the variable is x, and that the data line is #20. *&*&

n The axis of symmetry of a quadratic function is the line x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a).
*&*& The axis of symmetry of a quadratic function is the line x = -b / (2 a). *&*&

y The values of y which correspond to t = c are found by plugging c in for t and evaluating the function to find y.

n The values of y which correspond to t = c can be found by locating y = c on the vertical axis, moving horizontally along the line y = c to the graph, then moving vertically to the t axis.

y To author an expression you have to get into the narrow box near the bottom of the screen.

y You can plot a data set in DERIVE using the syntax [ [2, 9], [3, 12], [4, 18] ], for example, to represent y values 9, 12 and 18 vs. x values 2, 3, 4.

y If you plot a data set such as [ [2, 9], [3, 12], [4, 18] ], three points will appear on the graph, provided that the graph has an appropriate plot range.

n If you author an equation you can click on Plot and follow a few simple steps to get the solution.

n Every point of the graph of y = f ( x - h ) lies -h units in the horizontal direction from the corresponding point of the graph of y = f(x).

n If you plot the expression (x-2)(x+3) and the expression 2x + 7, DERIVE will automatically scale the screen so you can see where the two graphs cross.

*&*& DERIVE won't automatically scale in order to show you what you think you want to see. You gotta pick an appropriate scale and set the plot range. *&*&

n The data set [ [2, 9], [3, 12], [4, 18] ] would be represented in DERIVE as (2, 9), (3,12), (4,18).

n If you plot a data set such as [ [2, 9], [3, 12], [4, 18] ], three points and a best-fit line will appear on the graph, provided that the graph has an appropriate plot range.

Correct answer	Question
n	The zeros of the function y = f(t) are found by plugging 0 in for t and evaluating the function to find y. && You find the zeros by setting f(t) equal to 0 and solving for t. &&
y	There are parabolic functions for which the vertex is the lowest point.
y	If the graph of a quadratic function passes through the horizontal axis it does so at horizontal coordinates (-b +- sqrt(b^2 - 4 a c) ) / (2a).
y	For a function y = f(t), there can be only one value of y which corresponds to t = 0. && Plug in t = 0 you get y = f(0), whatever that equals. That's all you get. We're implicitly assuming that 0 is in the domain of the function, and that should be stated in the problem statement. Don't confuse with the fact that a function can have many zeros. That's where y = 0, not where t = 0. &&
n	For a function y = f(t), there can be any number of values of y which correspond to t = 0. && Not so. There is exactly one value, assuming that 0 is in the domain of the function. &&
y	The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis. If t is a valid value of the independent variable (i.e., if t is in the domain of the function) then there will be exactly one such point.
n	The values of t for which the function y = f(t) is equal to c are found by evaluating the expression f(c). && y = f(t) is equal to c if y = c, not if t = c. &&
n	The equation f(t) = c might have any finite number of solutions but cannot have a infinite number. && for example a sine-wave type function can take any y value in its range and infinite number of times. &&
n	The values of y which correspond to t = 0 are represented by the points where the graph of the function crosses the t axis. && t = 0 points are on the y axis, not necessarily the t axis. A t = 0 point is your y=intercept. &&
y	The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis.
y	The values of y which correspond to t = c can be found by locating t = c on the horizontal axis, moving upward along the line t = c to the graph, then across to the y axis. If t is a valid value of the independent variable (i.e., if t is in the domain of the function) then there will be exactly one such point.
n	If a data set such as [ [2, 9], [3, 12], [4, 18] ] is in line #20, then the command fit( [x, a x^2 + b x + c ] , #20) can be simplified to give you the equation of the parabola that best fits the data. && fit( [x, a x^2 + b x + c ] , #20) tells Derive that the form is a x^2 = b x + c and the variable is x, and that the data line is #20. &&
n	The axis of symmetry of a quadratic function is the line x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a). && The axis of symmetry of a quadratic function is the line x = -b / (2 a). &&
y	The values of y which correspond to t = c are found by plugging c in for t and evaluating the function to find y.
n	The values of y which correspond to t = c can be found by locating y = c on the vertical axis, moving horizontally along the line y = c to the graph, then moving vertically to the t axis.
y	To author an expression you have to get into the narrow box near the bottom of the screen.
y	You can plot a data set in DERIVE using the syntax [ [2, 9], [3, 12], [4, 18] ], for example, to represent y values 9, 12 and 18 vs. x values 2, 3, 4.
y	If you plot a data set such as [ [2, 9], [3, 12], [4, 18] ], three points will appear on the graph, provided that the graph has an appropriate plot range.
n	If you author an equation you can click on Plot and follow a few simple steps to get the solution.
n	Every point of the graph of y = f ( x - h ) lies -h units in the horizontal direction from the corresponding point of the graph of y = f(x).
n	If you plot the expression (x-2)(x+3) and the expression 2x + 7, DERIVE will automatically scale the screen so you can see where the two graphs cross. && DERIVE won't automatically scale in order to show you what you think you want to see. You gotta pick an appropriate scale and set the plot range. &&
n	The data set [ [2, 9], [3, 12], [4, 18] ] would be represented in DERIVE as (2, 9), (3,12), (4,18).
n	If you plot a data set such as [ [2, 9], [3, 12], [4, 18] ], three points and a best-fit line will appear on the graph, provided that the graph has an appropriate plot range.