Hot X: Algebra Exposed

Transcription

1 Hot X: Algebra Exposed Solution Guide for Chapter 11 Here are the solutions for the Doing the Math exercises in Hot X: Algebra Exposed! DTM from p Since m = 2, our equation will look like this: y = 2x + b. To find b, we ll plug in the point (, ) and see what b would have to be in order for the statement to be true: y = 2x + b = 2() + b = 6 + b b =. This means our equation is y = 2x. Done! Answer: y = 2x. Since m =, our equation will look like this: y = x + b. To find b, we ll plug in the point (1, 2) and see what b would have to be in order for the statement to be true: y = x + b 2 = (1) + b 2 = + b b = 1. This means our equation is y = x 1. Done! Answer: y = x 1 4. Since m = 1, our equation will look like this: y = x + b. To find b, we ll plug in the point (0, 0) and see what b would have to be in order for the statement to be true: y = x + b 0 = (0) + b 0 = 0 + b b = 0. This means our equation is y = x + 0; in other words, y = x. Done! Answer: y = x

2 5. Since m = 1 4, our equation will look like this: y = 1 x + b. To find b, we ll plug in the 4 point (8, 6) and see what b would have to be in order for the statement to be true: y = 1 4 x + b 6 = 1 (8) + b 6 = 2 + b b = 4. 4 This means our equation is y = 1 4 x + 4. Done! Answer: y = 1 4 x + 4 DTM from p So what s the slope of the line we re given? The equation is in slope-intercept form, so we can see that its slope is. Since we want a line parallel to that line, our new line s slope will be, too! That means our semi-mysterious equation is y = x + b. What s b? Let s plug in the point we re given, (0, 0), and find out: y = x + b 0 = (0) + b 0 = b. So our line is y = x. To graph this, we can plot the y-intercept (0, 0), and then let s pick two other points. If we plug in x = 1, we get: y = x y = ( 1) y =. So we get the point ( 1, ). And if we plug in x = 1, we get: y = x y = (1) y =. So we get the point (1, ). To graph the original line, y = x +, we can plot its y-intercept, (0, ), and let s find two more points. Plugging in x = 1, we get: y = x + y = ( 1) + y = + y = 0. So we get the point ( 1, 0). Hey, that s the x-intercept! And if we plug in x = 2, we get: y = x + y = ( 2) + y = 6 + y =. So we get the point ( 2, ). Now we plot em and put lines through em. Yep, they look parallel!

3 Answer: y = x (and see graph below). This is almost like the same problem as above, except this time we want the new line to be perpendicular to y = x +. That means our new line s slope will be the negative reciprocal of, which is 1. That means our semi-mysterious equation looks like this: y = 1 x + b. Since this line is supposed to go through the origin, that means the y-intercept will be 0, which means b = 0. So our new line s equation is: y = 1 x. To graph this, we can use the y-intercept (0, 0), and let s find two more points. Let s plug in x = (to get rid of that fraction) and we get: y = 1 x y = 1 () y = 1. So we get the point (, 1). Next let s try plugging in x =, and we get: y = 1 x y = 1 ( ) y = 1. So we get the point (, 1). Now we can graph this new line, y = 1 x.

4 And we already found points and graphed the line y = x + in the previous problem, so we ll use those same points again. And yep, our two lines look perpendicular! Answer: y = 1 x (and see graph below) 4. This one is a bit strange just follow along! So, our new line is supposed to be perpendicular to the line y = 6, which has a slope of zero. But the negative reciprocal of zero is undefined! Recall from p.154 how vertical lines indeed have undefined slopes, so our new line will be vertical. an those have the form x = number. So what number should it be? Well, it needs to go through the point (2, 4), and in that point, the x-value is 2. So how about the equation of the line x = 2? Well gosh, we know these lines are perpendicular, and we know our new line goes through (2, 4), so we re done! Well, let s graph these two lines and see if it makes sense. The line y = 6 is the line that has points like (0, 6), (5, 6), ( 9, 6), etc. And the line x = 2 has points like (2, 0), (2, 4), (2, 12), etc. Answer: x = 2 (and see graph below)

5 5. So our new line will be parallel to 2x + y = 5. So what s its slope? Let s put it into slope-intercept form to find out. Subtracting 2x from both sides and then dividing both sides by, we get: y = 2 x + 5, so we can see its slope is 2. Our new line s slope will be the same! So our new line will look like: y = 2 x + b. To find b, we ll plug in the point we were given: (, 1), and we get: y = 2 x + b 1 = 2 () + b 1 = 2 + b b = So our new line s equation is: y = 2 x + To graph this, we ll use the y-intercept, (0, ), and the point we re given, (, 1) and also find one other point. Let s plug in x = to get rid of the fraction, and we get: y = 2 x + y = 2 ( ) + y = 2 + y = 5. So we get a third point, (, 5), and we can graph our new line!

6 To graph the original line, 2x + y = 5, (in other words, y = 2 x + 5 ), let s find some points by looking for easy x-values to plug in. Now, most values of x will give fractions for their corresponding y-values, but I m going to pick ones that don t. (It just takes a little experience and cleverness, that s all!) Plugging in x = 1, we get: y = 2 x + 5 y = 2 (1) + 5 y = y = 5 2 y = y = 1. So we get the point (1, 1). Plugging in x = 4, we get: y = 2 x + 5 y = 2 (4) + 5 y = y = y = y = 1. So we get the point (4, 1). Finally, let s plug in x = 2, and we get: y = 2 x + 5 y = 2 ( 2) + 5 y = y = 9 y =. So we get the point ( 2, ), and we re ready to graph the line. Done! Answer: y = 2 x + (and see graph below)

7 6. This is the same original line as in the previous problem, 2x + y = 5, but this time we want our new line to be perpendicular to it. Recall that we found that its slope is 2. So, our new line s slope will be the negative reciprocal of that, which is 2. Great, so now we know our line will look like this: y = x + b. To find b, let s plug in 2 the point we ve been given, (, 1), and we get: y = 2 x + b 1 = 2 () + b 1 = b b = b = b = 7. So our new line s equation is: 2 y = 2 x 7 2. Time to find points to plot! We know it passes through (, 1). So let s find two others. Let s plug in x = 5, and we get: y = 2 x 7 2 y = 2 (5) 7 2 y = y = 8 2 y = 4. So we have the point (5, 4). Next let s plug in x = 1, and we get: y = 2 x 7 2 y = 2 (1) 7 2 y = y = 4 2 y = 2. So we get a third point, (1, 2), and we re ready to graph the new line! And we already graphed the original line, 2x + y = 5, in the previous problem, so we ll use those same points. Yep, our lines look perpendicular. Done! Answer: y = 2 x 7 2 (and see graph below)

8 7. Okay, we re given the line y = 0.25x + 1. We want a new line, perpendicular to this one. Since the slope of this line is 0.25, we need to find the negative reciprocal of Well the negative part is easy, but what s the reciprocal of 0.25? Let s first write it as a fraction! You probably know that 0.25 = 1, but let s say that you forgot. Let s review 4 how to do this. So, to write 0.25 as a fraction, first we can do this: 0.25 = , right? Then let s use a copycat, 100, to get rid of those pesky decimal places, and because it s a 100 copycat, it doesn t change anyone s value: this: = = 1 4. = = 25, which reduces like 100 Okay, now we re ready to find the reciprocal, which of course is just 4! So we have our new line s slope: 4, which we just saw is the negative reciprocal of That means our new line will look like this: y = 4x + b. To find b, we ll plug in the point that was given to us: (1, 1), and we get:

9 1 = 4(1) + b 1 = 4 + b b =. So our new line is: y = 4x. To graph this, we can use the point (1, 1), and also the y-intercept, which we can see from the equation, is (0, ). For our third point, let s plug in x = 2 and we get: y = 4x y = 4(2) y = 8 y = 5. So we get the point (2, 5). And we re ready to graph this line! To graph the original line, y = 0.25x + 1, let s pick some points that make those decimal places go away. Actually, x = 4 will do the trick, since: 0.25(4) is the same as 1 4 (4), which is just 1! So plugging in x = 4, we get: y = 0.25x + 1 y = 0.25(4) + 1 y = y = 0. So we get the point (4, 0). Next let s pick the value x = 4, and we get: y = 0.25x + 1 y = 0.25( 4) + 1 y = y = 2. So we get the point ( 4, 2). And now we can graph both lines. They sure look perpendicular to me! Answer: y = 4x (and see graph below)

10 DTM from p Our first job is to find the slope of the line connecting these two points. Okay, so first let s pick a favorite point, how about (, 5)? So that means we re saying that x 1 =, y 1 = 5, x 2 = 2, and y 2 = 4. Plugging that into our slope equation, we get: m = y 1 y 2 m = 5 4 x 1 x 2 2 m = 1 1 m = 1. Great! So now we know our line s equation will look something like this: y = 1x + b y = x + b. To find b, let s use one of our points; how about our favorite point, (, 5), and we get: y = x + b 5 = + b b = 2 That means the equation of the line connecting our two points is y = x + 2. Done! Answer: m = 1; y = x + 2. Okay, so to find the slope, let s pick a favorite point, how about (0, 1)? Then: that x 1 = 0, y 1 = 1, x 2 = 2, and y 2 = 0, and we get: m = y 1 y 2 x 1 x 2 m = m = 1 2 m = 1 2 Now we know the line s equation will take the form: y = 1 2 x + b To find b, let s plug in one of our points. Let s use (2, 0): y = 1 2 x + b 0 = 1 (2) + b 0 = 1 + b b = 1 2 So our equation is: y = 1 2 x + 1 Answer: m = 1 2 ; y = 1 2 x + 1

11 4. I like zeros, so let s make the favorite point (0, 5), and that means: that x 1 = 0, y 1 = 5, x 2 =, and y 2 = 7. Plugging them into our slope formula, we get: m = y 1 y 2 x 1 x 2 m = m = 2 m = 2 Great, now we know the equation will look like this: y = 2 x + b. To find b, let s plug in the point (0, 5) and we get: y = 2 x + b 5 = 2 (0) + b 5 = 0 + b b = 5 And our equation is: y = 2 x + 5 Answer: m = 2 ; y = 2 x This time, let s pick (2, 2) as our favorite point. That means: x 1 = 2, y 1 = 2, x 2 = 0, and y 2 = 5. Plugging in these values to our slope formula, we get: m = y y 1 2 m = 2 5 x 1 x m = 7 2. Okay, so we know our equation will take the form: y = 7 x + b. Let s find b by 2 plugging in the point (0, 5), and we get: y = 7 2 x + b 5 = 7 2 (0) + b 5 = 0 + b b = 5 So our full equation is: y = 7 2 x + 5. Done! Answer: m = 7 2 ; y = 7 2 x + 5