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Find the equations of the lines and plot the lines from Exercise $56$.

$y=-5$

Algebra

Chapter 1

Functions and their Applications

Section 1

The Line

Functions

Oregon State University

Harvey Mudd College

Idaho State University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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01:55

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Equations of Lines Find an…

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this problem is referencing exercise number 56 from earlier in this, um, a section. So let's go back and look at 56 for a moment. In at number 56 we were given two points, one negative five and to negative five. And in that exercise, we found the slope of the line connecting them, which was zero. So, given that information for this problem, we want to find the equation of the line that all of this represents and then graph it. So what we have here? We have points and we have a slope. We don't have the y intercept. So I don't want to use the, uh, slope intercept form y equals MX plus B instead. Since I have a point and the slope, we use the point slope form. Why? Minus why one equals m times X minus X one. So x one y one is a point that I know, and I could pick either one of these points. It actually doesn't matter that both could be the same answer. And my slope, we'll go right there. So let's plug in our values. I'm just gonna take the first point just because it's first. Why minus why once So subtracting negative five is the same is adding five. My slope is zero X minus one. Well, zero types. Anything. This whole right hand side is going to be zero, which means I'm gonna end up with y equals negative five. So let's graph that y equals negative five. First, I'm going to put 1234 years negative five of my y axis. If you notice there is no X value in this equation, that means it doesn't matter what the X value is. I could pick any X I want as long as I maintain the same y value. Why equals negative five and you can see that in the points we were given the exchanges, but the Y value remains the same, so I could have one negative five to negative five. I put those those blue dots on the line there. Yeah, I could have six negative five negative eight negative five, which is somewhere right about over there. I didn't quite count that X doesn't matter as long as the y equals negative five. So when you see why, equaling a constant, you know that's going to be horizontal line. So in this case, this horizontal line is why equals negative five

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