Question:  `q001.  Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis.

• Plot the points (1, -2), (3, 5) and (7, 8). 

• Sketch a smooth curve passing through these three points. 

On your curve, what are the y coordinates corresponding to x coordinates 1, 3, 5 and 7?  Estimate these coordinates as accurately as you can from your graph. 

Retain your sketch for use in future questions. 

Your solution: 

Confidence rating:

Given Solution: 

The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 5 and 8, respectively, of those points. 

• At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. 

• You are unlikely to have drawn a perfect parabola and your estimate will almost certainly differ from this value.  Any estimate with .5 or so of this value would be a good estimate. 

Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7.   

• The peak of the actual parabola will occur close to x = 6.  Any estimate between x = 5 and x = 7 is pretty good, and even an estimate between x = 7 and x = 8 is not unreasonable.

Self-critique (if necessary):

Self-critique Rating:

Question:  `q002.  Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7.  Also estimate the x values at which y is 0.

Your solution: 

Confidence rating:

Given Solution: 

The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7.  You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. 

For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10).

• y = 3 near x = 2.3 (and near x = 9.3). 

• y = 5 at the given point (3, 5), where x = 3. 

• y = 7 near x = 4 (and also near x = 7.7).

Any set of estimates in the vicinity of these values is good.

STUDENT COMMENT:

I’m still not clear on this. I didn’t get the same as you and my graph went to 12. I know when we estimate we will not get
the same answer, but did I do this right?  If not what do I do to correct it?

INSTRUCTOR RESPONSE: 

The following commentary need not be self-critiqued, though you may certainly ask questions if you wish.  However I have two reasons for including it:

• The estimates here were given by a good student whose graph for some reason didn't match the points on this particular problem.  If this student missed his/her estimates, then most students will need some clarification at this point.

• The discussion and the graphs shown here should be helpful to you.  You should be able to sketch these graphs, label the coordinate axes based on the coordinates of the three points, and make reasonable verification of the values given here.

• This situation is closely related to the work we will be doing in the first few assignments, where we will actually learn to do parabolic or quadratic curve fits.

Instructor response to student:

Some of your estimates were reasonable, but others weren't.

Your solution was

if y = 0 then x = 0
if y = 1 then x = approx. 1.7
if y = 3 then x = approx. 4.6
if y = 5 then x = approx. 8.1
if y = 7 then x = approx. 11.5

If the x coordinates 1,3 7 match the y coordinates -2,5,8, then

y = 1 at x = 1.7 is a very reasonable estimate

y = 3 at x = 4.6 is not a reasonable estimate. By the time x = 3, y already has a value of 5, and is increasing toward value y = 8 at x = 7. For an x value between 3 and 5 we would expect y to be greater than 5; we would not expect a y value of 3 to occur on this x interval.

y = 5 at x = 8.1 is a reasonable estimate. When x = 7 we have y = 8, and it's possible to draw a reasonable graph on which y decreases to 5 by the time x = 8.

y = 7 at x = 11.5 isn't consistent with your previous estimates. You already have y down to value 5 by the time x = 8.1, and there's no reason to expect the graph to turn around and go back up to 7 by the time x reaches 11.5.
 

The figure below depicts the three data points.  Your graph should have similar form.

The graph below shows the parabola that actually passes through these points.  Your graph won't necessarily look the same, since you probably aren't practiced at drawing an accurate parabola through three given points, and besides nobody told you to use a parabola.  So it's no problem if your graph differs from this one, but this 'ideal' graph is the basis for the numbers in the given solution.  For this graph, y = 0 somewhere between x = 1 and x = 3, closer to x = 1; we might estimate that this occurs at x = 1.5 (a fairly precise value for the actual parabola is about x = 1.476). 

You will learn a lot more about parabolas within the next couple of weeks.  In fact, by Assignment 4 or 5, you will have learned how to create and use a parabolic model for any three points, and you'll see that while it's not really what you'd call easy, it's not all that difficult either.

The graph below isn't parabolic, but it's still a smooth curve and passes through the given points, and your graph could well look like this. 

The y value at x = 5 appears to be pretty close to the x = 7 value, which is y = 8.  The y = 0 value again occurs between x = 1 and x = 3; perhaps a little to the left of the point in the above graph, maybe at x = 1.4.

Another reasonable graph is shown below.  At x = 5 the y value is probably around 7.

The graph below agrees with the estimates you gave.  While it fits the overall trend of the data by 'staying in the middle', it doesn't fit any of the three given points very well. 

This graph has been included so you can verify that it agrees reasonably with your given values

if y = 0 then x = 0
if y = 1 then x = approx. 1.7
if y = 3 then x = approx. 4.6
if y = 5 then x = approx. 8.1
if y = 7 then x = approx. 11.5

Self-critique (if necessary):

Self-critique Rating:

Question:  `q003.  Suppose the graph you used in the preceding two exercises represents the profit y on an item, with profit given in cents, when the selling price is x, with selling price in dollars.  According to your graph:

• What would be the profit if the item is sold for 4 dollars? 

• What selling price would result in a profit of 7 cents? 

Your solution: 

Confidence rating:

Given Solution: 

To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph. 

• This point is easily located by sketching a vertical line through x = 4.  Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 6 of 7, depending on how you drew the graph.  This represents a profit of about 6 or 7 cents.

The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 or 5 (y = 7 would also occur at x = 7.7, approx., if the entire parabola was drawn).

STUDENT COMMENT:

I think I'm doing my graphs wrong.  Can you help?

INSTRUCTOR RESPONSE:

See also my notes on the preceding problems, especially the ones with the graphs.

The graph depicted below doesn't fit the original points very well, but we'll use it to illustrate the situation.  As before you should review the coordinates of the three given points and make sense of them on this graph.

The 'red' line is a vertical line through x = 4.  This line is drawn first.

The 'green' horizontal line is then drawn through the point where the x = 4 line intersects the graph.

You should verify for yourself that the 'green' line intersects the y axis around y = 6 (the actual coordinate is about y = 6.2, but you aren't expected to be able to verify the value that accurately; you should however be able to tell that the coordinate is closer to 6 than to 5 or 7, based on the coordinates of the three given points).

This would tell us that if the selling price is around 4 units, the profit is around 6 units.

To find the selling price that will bring a profit of 7 units, we first draw the horizontal 'blue' line through y = 7.

Then we draw the vertical 'purple' line, through the point where our horizontal line intersects the graph.  This line intersects the x axis somewhere between x = 4 and x = 5.  It's hard to tell whether the x value is closer to 4 than to 5, so we might estimate that this occurs around x = 4.5 (the actual value as depicted is about x = 4.43).

This would tell us that to achieve a profit of 7 units, the selling price should be about 4.5 units.

Your estimates will of course differ from these, depending on how you drew your graph. 

For example, if the parabola is taken as the 'perfect' model for the data, the results might vary from these by as much as a unit.

Worth noting, even though you probably don't understand it yet:  For a number of reasons you will soon understand, it turns out that the parabola is a very reasonable model for estimating how price and profit are related.

Self-critique (if necessary):

Self-critique Rating:

Question:  `q004.  On another set of coordinate axes, plot the points (-3, 4) and (5, -2).  Sketch a straight line through these points. 

We will proceed step-by-step obtain an approximate equation for this line:

First substitute the x and y coordinates of the first point into the form y = m x + b. 

• What equation do you obtain when you make this substitution?

Your solution: 

Confidence rating:

Given Solution: 

Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation

•  4 = -3 m + b. 

We can reverse the right- and left-hand sides to get

• -3 m + b = 4. 

Self-critique (if necessary):

Self-critique Rating:

Question:  `q005.  Substitute the coordinates of the point (5, -2) into the form y = m x + b.  What equation do you get?

Your solution: 

Confidence rating:

Given Solution: 

Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation

•  -2 = 5 m + b. 

Reversing the sides we have

• 5 m + b = -2

Self-critique (if necessary):

Self-critique Rating:

Question:  `q006.  You have obtained the equations -3 m + b = 4 and 5 m + b = -2.  Use the method of elimination to solve these simultaneous equations for m and b.

Your solution: 

Confidence rating:

Given Solution: 

Starting with the system

-3 m + b = 4

5 m + b = -2

we can easily eliminate b by subtracting the equations.  If we subtract the first equation from the second we obtain

-8 m = 6, with solution

m = -3/4.

Substituting this value into the first equation we obtain

(-3/4) * -3 + b = 4, which we easily solve to obtain

b = 7/4.

To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining

5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2.

Self-critique (if necessary):

Self-critique Rating:

Question:  `q007.  Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. 

• What equation do you obtain? 

• What is the significance of this equation?

Your solution: 

Confidence rating:

Given Solution: 

Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation

y = -3/4 x + 7/4.

This is the equation of the straight line through the given points (-3, 4) and (5, -2).

Self-critique (if necessary):

Self-critique Rating:

Question:  `q008.  The vertex of a parabola is at (4, -6), and the point (3, -3) is also on the parabola. Give the coordinates of one other point on the parabola.

Your solution: 

Confidence rating:

Question:  `q009.  If

2 x + 7 y = 8

and

4 x - 3 y = -18

then what is the value of x? What therefore is the value of y?

Your solution: 

Confidence rating:

Question:  `q010.  Sketch a smooth curve through the points (2, -1), (3, -2), and (5, 2). According to your sketch, what is your best estimate of the value of x when the graph passes through the x axis?

Your solution: 

Confidence rating:

Self-critique rating:

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