Math 1: Solutions to Written Homework 1 Due Friday, October 3, 2008

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1 Instructions: You are encouraged to work out solutions to these problems in groups! Discuss the problems with your classmates, the tutors and/or the instructors. After working doing so, please write up your solutions legibly on a separate sheet (or sheets) of paper, showing all of your work (this part should be done on your own). When you are asked to give explanations, be sure to use complete sentences. You are welcome (and sometimes encouraged) to use calculators or computing devices. 1. For a quadratic equation, y = ax 2 + bx + c, the discriminant is = b 2 4ac (the part that goes under the square root in the quadratic formula). Find an example of a quadratic equation with discriminant 0 where none of the coefficients are 0 (so a 0, b 0 and c 0) and sketch it s graph. Solution: There are many possible examples, here are a few: y = x 2 +2x+1, y = x 2 2x+1, y = x 2 + 4x + 4, y = 4x 2 + 4x + 1, etc. Examples need not have integer coefficients. We say that a quadratic equation has a double root when = 0. Explain (in complete sentences) what this means for the graph and why we might call it a double root. Hint: For the first part of this question, draw a few examples and see what the graphs have in common, then try to make sense of this using what you know about the quadratic formula. For the second part, again consider the quadratic formula - we expect it to give us two roots of a quadratic equation; what happens when = 0? Solution: When = 0, the graph of the quadratic equation just touches the x-axis (that is, it does not cross it). When we use the quadratic formula to solve for the roots of y = ax 2 + bx + c, where = b 2 4ac = 0, we only find a single solution: x = b. This 2a means we can rewrite the equation as y = ax 2 + bx + c = a(x b 2a )2, hence b is not just 2a a root, but a double root. Writing it in this form, y = a(x b 2a )2 also makes it more clear why the graph never crosses the x-axis: since the quantity (x b 2a )2 is a squared quantity, it will always be 0, so if a > 0 we ll always have y 0 (and likewise if a < 0, we ll always have y 0), hence the graph never crosses the x-axis. Now find all values of b for which x 2 bx + 1 = 0 has a double root. Solution: For x 2 bx + 1 = 0 to have a double root, we must have = ( b) = 0 or b 2 4 = 0. Hence b 2 = 4 and b = ±2. 2. Problem 16 from Section 1.1 in the book: You place a frozen pie in an oven and bake it for an hour. Then you take it out and let it cool before eating it. Describe how the temperature of the pie changes as time passes. Then sketch a rough graph of the temperature of the pie as a function of time. Solution: The temperature of the pie would start low (say around 30 F) and rise to something like 375 F, where it would remain constant until removed from the oven (so the whole rising and staying constant would take 1 hour). After you remove it from the oven (at 1 1

2 hour) the temperature would start to decline, but not quite as sharply as it had risen (since the difference between room temperature and the temperature of the oven is not as great as between frozen and the oven temperature). If you let it cool to room temperature before eating it, it will decrease to somewhere around 70 F and level off. 3. Calculators are encouraged on this problem to avoid rounding errors. The goal of this problem is to find a line that touches the graph of y = x 2 + 2x + 1 = (x + 1)(x + 1) at (0, 1) and nowhere else. This line is called the tangent line to y = x 2 + 2x + 1 at the point (0, 1). First let s warm up by looking at the point ( 1, 0) on y = x 2 + 2x + 1. Consider the line y = 0. This line appears to touch y = x 2 + 2x + 1 at the point ( 1, 0) and nowhere else. This is in fact true. First, it does touch y = x 2 + 2x + 1 at ( 1, 0) because this point is a solution for both y = x 2 + 2x + 1 and y = 0. (a) Verify that the point ( 1, 0) is indeed a solution to y = x 2 + 2x + 1 and y = 0. Show all your work. Solution: Since the y-coordinate of ( 1, 0) is 0, we know it lies on the line y = 0. To see that it is a solution to y = x 2 + 2x + 1, we will simply plug in 1 for x and see that we get 0 for y: ( 1) 2 + 2( 1) + 1 = 1 + ( 2) + 1 = 0. Now let s see if there are any more places where y = x 2 + 2x + 1 and y = 0 touch. Notice that the graph suggests that there aren t. Indeed, any other place where y = x 2 + 2x + 1 touched y = 0 would be a solution of 0 = x 2 + 2x + 1, and as the quadratic formula tells us 1 is the only solution to this equation. Let s move on to the main problem. We will break it into several parts. (b) Find the equation of the line that goes through (0, 1) and (1, 4). Note that BOTH 2

3 of these points are on the graph of y = x 2 + 2x + 1. Solution: Since one of the points we are given to work with is (0, 1), we know the y-intercept of the desired line will be 1, so all we need to do is find the slope of the line and we ll have everything we need to write an equation in slope-intercept form. Recall that the slope of a line can be found by taking the quotient y 2 y 1 x 2 x 1, where (x 1, y 1 ) and (x 2, y 2 ) are any two points on the line. Hence, we have: = 3 1 = 3 And so an equation for our line is: y = 3x + 1 (c) Show your work in checking that the two points (0, 1) and (1, 4) are on the line you found in part (b). Solution: As before, we will verify this by pluggin in our x values and seeing that the correct y values result: = = 4 (d) Explain why these are the only two points where your line from part (b) intersects the graph of y = x 2 + 2x + 1. Solution: To find the points where the line y = 3x+1 intersects the graph of y = x 2 +2x+1, we set the two equations equal to one another: 3x + 1 = x 2 + 2x + 1. We can simplify this to: 0 = x 2 x = x(x 1). Hence the only two points of intersection occur when x = 0 and when x = 1. (e) Find the equation of the line that goes through through (0, 1) and (0.5, 2.25). Note = = 2.5 y = 2.5x + 1 (f) Show your work in checking that the two points (0, 1) and (0.5, 2.25) are on the line you found in part (e). Solution: Once again, we do this by plugging in the values 0 and 0.5 for x and verifying 3

4 = = = 2.25 (g) Find the equation of the line that goes through through (0, 1) and (0.1, 1.21). Note = = 2.1 y = 2.1x + 1 (h) Show your work in checking that the two points (0, 1) and (0.1, 1.21) are on the line you found in part (g). Solution: Once again, we do this by plugging in the values 0 and 0.1 for x and verifying = = = 1.21 (i) Find the equation of the line that goes through through (0, 1) and (0.01, ). Note = = 2.01 y = 2.01x + 1 (j) Show your work in checking that the two points (0, 1) and (0.01, ) are on the line you found in part (i). Solution: Once again, we do this by plugging in the values 0 and 0.01 for x and verifying 4

5 = = = (k) Graph the lines you found in parts (b), (e), (g) and (i) on a single graph. Solution: The graph should clearly show the secant lines getting closer and closer to the desired tangent line. (l) What do you think the equation of the tangent line to the graph y = x 2 + 2x + 1 at the point (0, 1) is? What did you see in parts (b) through (j) that would suggest this? Solution: Since the slopes seem to be getting closer and closer to 2, the equation for the tangent line is probably y = 2x