Question

If you trace along the graph of $y=1.5 x-6$ using an integer window, which of the following coordinates would not be displayed? a. $(0,-6)$ b. $(1,-4.5)$ c. $(4,0)$ d. $(1.2,-4.2)$

    If you trace along the graph of $y=1.5 x-6$ using an integer window, which of the following coordinates would not be displayed?
a. $(0,-6)$
b. $(1,-4.5)$
c. $(4,0)$
d. $(1.2,-4.2)$
Intermediate Algebra: A Graphing Approach
Intermediate Algebra: A Graphing Approach
Elayn Martin-Gay,… 5th Edition
Chapter 2, Problem 43 ↓
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If you trace along the graph of $y=1.5 x-6$ using an integer window, which of the following coordinates would not be displayed? a. $(0,-6)$ b. $(1,-4.5)$ c. $(4,0)$ d. $(1.2,-4.2)$
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The graph of $f(x)$ is shown above in the $x y$ -plane. The points $(0,3),(5 b, b),$ and $(10 b,-b)$ are on the line described by $f(x)$ . If $b$ is a positive constant, what are the coordinates of point $C ?$ A) $(5,1)$ B) $(10,-1)$ C) $(15,-0.5)$ D) $(20,-2)$


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Transcript

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00:01 All right, so the first thing we need to do is calculate our slope.
00:05 The formula for a slope for this line would be y2 minus y1 over x2 minus x1.
00:12 All right, and you need to pick two points on your graph.
00:16 So the points that i picked were, i can write them over here, 5b and 10b negative b.
00:29 So i'm going to plug these two points into my formula to calculate my slope.
00:34 So y2 in this case is going to be negative b, right? that's our y2.
00:41 So negative b minus y1 is b.
00:45 So negative b minus me divided by x2.
00:49 X2 is 10b minus 5b minus that x1.
00:55 All right.
00:55 B minus negative b minus b is a negative 2b 10b minus 5 b is a 5b b divided by b is 1 so these are going to cancel out so we have a slope is negative 2 over 5 all right now the other problem or the other point that it gave us on the graph was the y intercept in the problem and it tells us the y intercept is 0 -3.
01:29 So if ever you have the slope and the y intercept, you can write your equation in point or slope intercept form.
01:36 And so we could write y equals our slope negative 2 over 5x plus 3.
01:43 That's just the slope intercept form of an equation.
01:46 You can do that if you have the slope and the y intercept, in which case we have both.
01:51 So this is the equation of the line.
01:53 Now it doesn't just ask us for the equation of the line, right? it wants to know what point c is.
01:59 In order for us to know point c, we need to solve for b.
02:05 So we need to plug in a plot point into our equation and solve for b...
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