describing graph

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Given Solution: `aThe graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `aThe graph forms a straight line with no change in steepness. STUDENT COMMENT Ok, I may not understand what exactly it meant by steepness, I was thinking since it was increasing it would also be getting steeper????? INSTRUCTOR RESPONSE A graph can increase while getting steeper and steeper; or it can increase while getting less and less steep. Or it can increase with no change in steepness. Analogies: When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep. When you go up a ramp the steepness stays the same until you get to the end of the ramp. When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `aBetween any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change. EXPANDED EXPLANATION Any student who has completed Algebra I and Algebra II should be familiar with slope calculations. Most students are. However a number of students appear to be very fuzzy on the concept, and I suspect that not all prerequisite courses cover this concept adequately (though I am confident that it's done well at VHCC). Also a number of students haven't taken a math course in awhile, and might simply be a bit rusty with this idea. In any case the following expanded explanation might be helpful to some students: Slope = rise / run. The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate. Our function is y = 3 x - 4. When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2. When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20. The graph therefore contains the points (2, 2) and (8, 20). You should have made a graph showing these points. If not you should do so now. As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18. Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6. The slope is rise / run = 18 / 6 = 3. The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3. STUDENT QUESTION Am I not allowed to utilize my calculus tools, yet? Couldn't I have just taken the derivative for the function, y = 3x -4 to obtain 3 as the slope? However, I do know how to do both ways. Which is the more preferred method? INSTRUCTOR RESPONSE This exercise develops a language for describing some aspects of graphs, and does not assume calculus tools. Of course it's fine to use the calculus tools if you have them, as long as you understand the problem at the more basic level as well. Unfortunately, not every student who has had a calculus course would know how to apply those tools to this situation (for example, I've had students from other institutions who have made A's inApplied Calculus courses from other (not particularly reputable) institutions, who didn't understand the concept of a slope). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&. STUDENT QUESTION: I am a little hazy on what the steepness is INSTRUCTOR RESPONSE: The hill analogy I used above might be helpful. Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope. Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3. NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope. Since in this case the slope is positive, which implies that the function is increasing, the increasing slope therefore implies that the value of the function is increasing at an increasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave upward. This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph on this interval is decreasing at a decreasing rate. NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope. Since in this case the slope is negative, which implies that the function is decreasing, the increasing slope therefore implies that the rate of decrease is decreasing. The value of the function is therefore decreasing at a decreasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as decreasing and concave upward. This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties. STUDENT QUESTION: I am still unsure why the steepness is decreasing, I see why going from right to left, but the graph looks linear? INSTRUCTOR RESPONSE: The y value increases, but it changes by less and less for every succeeding x value. So the graph is increasing, but by less and less with each step. It's increasing but at a decreasing rate. The graph does not look linear. If it does, then it's probably because your x and/or y axis is not scaled in equal increments. NOTE FOR CALCULUS-PREPARED STUDENTS (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 1 / (2 sqrt(x)), which is positive but decreasing, and which therefore implies a decreasing slope. Since in this case the slope is positive, which implies that the function is increasing, the decreasing slope therefore implies that the rate of increase is decreasing. The value of the function is therefore increasing at a decreasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave downward. This could also be explained in terms of the second derivative, y '' = -1 / (4 x^(3/2)), which is negative on this interval. The negative second derivative implies that the graph is concave down. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

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Given Solution: `a** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. ** STUDENT QUESTION I don’t understand how the graph decreases at a decreasing rate because it decreases by half every time. The ½ is constant. INSTRUCTOR RESPONSE The values decrease by a factor of 1/2 every time. That means each number would be multiplied by 1/2 to get the next. As a result the numbers we are halving keep decreasing. Half of 5 is 2.5; half of 2.5 is 1.25; half of 1.25 is .625. The decreases from one number to the next are respectively 2.5, 1.25 and .625. If the y values 5, 2.5, 1.25, .625 are placed at equal x intervals, it should be clear that the graph is decreasing at a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

@& Excellent.

Note, of course, that this is so only if the acceleration is positivel *@

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Given Solution: `a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. ** STUDENT COMMENT I don’t fully understand a distance vs. time graph. INSTRUCTOR RESPONSE If y represents the distance from you to the car and t represents the time in seconds since the car started out, then the graph of y vs. t is a graph of distance vs. clock time. The car is speeding up, so in any series of equal time intervals it moves further with each new interval. The distance it moves on an interval is represented by the difference between the y coordinates, so if it move further during an interval the 'rise' of the graph on that interval will be greater. If the intervals are equally spaced along the t axis, the result is an increasing graph with increasing slope. This is best understood by sketching the graph according to this description. STUDENT QUESTION I understand the clock time but could you give me some examples of numbers to sketch a graph. I am drawing a blank to how to make myself understand.????? INSTRUCTOR RESPONSE If the car's velocity for the first second averages 1 ft / sec, then in subsequent second 3 ft / sec, then 5 ft / sec, then 7 ft / sec, it will move 1 foot during the first second, 3 feet during the next, 5 feet during the next and 7 feet during the next. A graph of velocity vs. clock time would be a straight line, since the velocity increases by the same amount every second. However the positions of the car, as measured from the starting point, would be position 1 foot after 1 second position 4 feet after 2 seconds (the position changes by 3 feet, started this second at 1 ft, so the car ends up with position 4 feet) position 9 feet after 3 seconds (the position changes by 5 feet, started this second at 4 ft, so the car ends up with position 9 feet) position 16 feet after 4 seconds (the position changes by 7 feet, started this second at 9 ft, so the car ends up with position 16 feet) So the graph of position vs. clock time has positions 0, 1, 4, 9 and 16 feet after 0, 1, 2, 3 and 4 seconds, respectively. The position vs. clock time graph is therefore increasing at an increasing rate. Let me know if this doesn't answer your question. STUDENT QUESTION I still don’t totally understand why it would necessarily be increasing at an increasing rate. Couldn’t it be a decreasing or even a standard rate as I mentioned above? INSTRUCTOR RESPONSE If the car speeds up then its distance from its starting position increases at an increasing rate. Its speed might be increasing at an increasing, constant or decreasing rate, but not its position. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK

@& Good work and excellent insights, very well expressed. *@