12. model with mathematics refer to the graphic novel frame below for exercises a - b.
a. write an equation in slope - intercept form for the total cost of any number of tickets at 7 tickets for $5.
b. write an equation in slope - intercept form for the total cost of a wristband for all you can ride.
h.o.t. problems higher order thinking
13. persevere with problems the x - intercept is the x - coordinate of the point where a graph crosses the x - axis. what is the slope of a line that has a y - intercept but no x - intercept? explain.
14. reason abstractly write an equation of a line that does not have a y - intercept.
15. justify conclusions suppose the graph of a line has a negative slope and a positive y - intercept. through which quadrants does the line pass? justify your reasoning.

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Accepted Answer

### Part a (12a) # Explanation: ## Step1: Define variables and slope Let $ x $ be the number of tickets, $ y $ be the total cost. The cost for 7 tickets is $5, so the cost per ticket (slope $ m $) is $ \frac{5}{7} $. The y - intercept $ b $ is 0 (no fixed cost other than per - ticket cost). The slope - intercept form is $ y=mx + b $. ## Step2: Write the equation Substitute $ m=\frac{5}{7} $ and $ b = 0 $ into the slope - intercept form. So the equation is $ y=\frac{5}{7}x $. ### Part b (12b) # Explanation: ## Step1: Define variables and slope Let $ x $ be the number of rides (or any variable representing usage), $ y $ be the total cost. The wristband costs a fixed $25, so the slope $ m = 0 $ (no cost per ride) and the y - intercept $ b=25 $. The slope - intercept form is $ y = mx + b $. ## Step2: Write the equation Substitute $ m = 0 $ and $ b=25 $ into the slope - intercept form. So the equation is $ y=25 $ (or $ y = 0x+25 $). ### Problem 13 # Explanation: ## Step1: Analyze the line's characteristics A line with a y - intercept but no x - intercept is a horizontal line? No, wait. A line with a y - intercept (crosses the y - axis) but no x - intercept (never crosses the x - axis) is a horizontal line? No, a vertical line has an undefined slope and no y - intercept (except when $ x = 0 $, but a vertical line $ x=a,a eq0 $ has no y - intercept). Wait, a line with a y - intercept but no x - intercept: a horizontal line $ y = k,k eq0 $. The slope of a horizontal line is 0. Because for a horizontal line, the change in $ y $ (rise) is 0, and slope $ m=\frac{\text{rise}}{\text{run}}=\frac{0}{\text{run}} = 0 $ (where run is the change in $ x $). ## Step2: Explain A line with a y - intercept ($ b eq0 $) and no x - intercept means it never crosses the x - axis. Such a line is horizontal (parallel to the x - axis). The slope of a horizontal line is 0, since the difference in $ y $ - values (rise) between any two points on the line is 0, and slope is $ \frac{\text{rise}}{\text{run}} $, so $ m = 0 $. ### Problem 14 # Explanation: ## Step1: Recall the y - intercept condition A line has no y - intercept when it is a vertical line (parallel to the y - axis). A vertical line has the equation $ x=a $, where $ a eq0 $ (if $ a = 0 $, it is the y - axis itself, which has all y - values, so it has a y - intercept). For example, take $ a = 3 $. ## Step2: Write the equation An example of a line with no y - intercept is $ x = 3 $ (or any $ x=a,a eq0 $). The general form is $ x = c $, where $ c$ is a non - zero constant. ### Problem 15 # Explanation: ## Step1: Recall the slope - intercept form The slope - intercept form is $ y=mx + b $, where $ m $ is the slope and $ b $ is the y - intercept. Given $ m<0 $ (negative slope) and $ b>0 $ (positive y - intercept). ## Step2: Analyze the quadrants - When $ x = 0 $, $ y=b>0 $, so the line passes through $ (0,b) $ which is in Quadrant I (since $ x = 0,y>0 $ is on the y - axis, but as $ x $ increases from 0, $ y=mx + b $ with $ m<0 $ will decrease. - For $ x>0 $, since $ m<0 $, as $ x $ increases, $ y $ decreases. When $ x = 0 $, $ y=b>0 $. Let's find the x - intercept: set $ y = 0 $, then $ 0=mx + b\Rightarrow x=-\frac{b}{m} $. Since $ m<0 $ and $ b>0 $, $ -\frac{b}{m}>0 $ (negative divided by negative is positive). So the x - intercept is at $ (-\frac{b}{m},0) $ where $ -\frac{b}{m}>0 $. - So when $ x = 0 $, $ y>0 $ (Quadrant I or on the y - axis). When $ x>0 $, as $ x $ increases from 0, $ y $ decreases from $ b $ to 0 (passing through Quadrant I when $ x $ is small and $ y>0 $, then crossing the x - axis at $ x =-\frac{b}{m}>0 $). When $ x<0 $, since $ m<0 $ and $ x<0 $, $ mx>0 $ (negative times negative is positive), so $ y=mx + b>0 + b>0 $. Wait, no: if $ x<0 $ and $ m<0 $, $ mx>0 $ (because negative times negative is positive), so $ y=mx + b>0 + b>0 $. But also, when $ x $ is a large negative number, $ y=mx + b $, since $ m<0 $, $ mx $ is a large positive number, and $ y$ will be positive. But we also have to consider the direction of the line. A line with negative slope goes down from left to right. - The line passes through Quadrant II (when $ x<0,y>0 $), Quadrant I (when $ 00 $), and Quadrant IV (when $ x>-\frac{b}{m},y<0 $). Wait, let's take an example: $ y=-x + 2 $. When $ x = 0 $, $ y = 2 $ (Quadrant I). When $ x = 2 $, $ y=0 $ (x - intercept). When $ x=3 $, $ y=-1 $ (Quadrant IV). When $ x=-1 $, $ y = 3 $ (Quadrant II). So the line passes through Quadrants II, I, and IV. - Justification: The positive y - intercept means the line crosses the y - axis above the origin (so it is in Quadrant I or on the y - axis at $ x = 0 $). The negative slope means the line decreases as $ x $ increases. So as $ x $ moves from negative to positive: for $ x<0 $, $ y=mx + b $, since $ m<0 $ and $ x<0 $, $ mx>0 $, so $ y>b>0 $ (Quadrant II). For $ 00 $ (Quadrant I). For $ x>-\frac{b}{m} $, $ y<0 $ (Quadrant IV). # Answers: 12a: $ \boldsymbol{y=\frac{5}{7}x} $ 12b: $ \boldsymbol{y = 25} $ (or $ \boldsymbol{y=0x + 25} $) 13: The slope is 0. A line with a y - intercept but no x - intercept is horizontal ($ y = k,k eq0 $), and horizontal lines have a slope of 0. 14: An example is $ \boldsymbol{x = 3} $ (or any $ \boldsymbol{x=c,c eq0} $) 15: The line passes through Quadrants II, I, and IV. The positive y - intercept ($ b>0 $) means it crosses the y - axis above the origin. The negative slope means it decreases as $ x $ increases. It crosses the x - axis at a positive $ x $ - value ($ x=-\frac{b}{m}>0 $ since $ m<0,b>0 $). For $ x<0 $, $ y>0 $ (Quadrant II); for $ 00 $ (Quadrant I); for $ x>-\frac{b}{m} $, $ y<0 $ (Quadrant IV).

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QUESTION IMAGE

Question

Explanation:

Part a (12a)

Step1: Define variables and slope

Step2: Write the equation

Part b (12b)

Step1: Define variables and slope

Step2: Write the equation

Problem 13

Step1: Analyze the line's characteristics

Step2: Explain

Problem 14

Answer: