Trendline with date axis - Microsoft Community
Graph your problem using the following steps: Type in your equation like y=2x+1 (If you have a second equation use a semicolon like y=2x+1 ; y=x+3); Press. Date: 01/08/97 at From: Lauren Subject: Writing Linear Equations in Point-Slope: (y-y1) = m(x-x1) y-y1 = mx - mx1 distributive property y = mx - mx1 + y1 which is the formula for calculating the y-intercept of a line from any point on the . If you would like further explanation on some point, or a graph to make this . This procedure does not work for finding the y-intercept, to do this she uses " value" in the CALC menu and enters X=0 to determine the y-value.
Divide both sides by 3. We get 3 is equal to x or x is equal to 3. So the point y is equal to 0, x is equal to 3 is on this line. And let me put it in order.
Graphing using intercepts (old) (video) | Khan Academy
So it's 3 comma 0. So this is the origin. That is 3 comma 0. This right here is the x-intercept. And remember, notice that point lies on the x-axis, but the y-value is 0. We haven't moved up or down. When you think x-intercept, you say, OK, that means my y-value is 0. So I have to solve for the x-value.
Now we do the opposite for the y-intercept.
And the y-intercept, we're sitting on this line, x-value must be 0. So let's figure out what y is equal to when x is equal to 0. So y is equal to-- I want to do it in that pink color. Well, 3 times 0 is just 0.
So 0 minus 9. Well that's, just equal to negative 9. So we have the point 0 comma negative 9. So when x is 0, we go down 9 for y. So right there is the point 0 comma negative 9.
Notice, it sits on the y-axis. That's why it's the y-intercept.
And the x-value is 0. We haven't moved to the left or right.How To Find The X and Y Intercepts of a Line
All you need is two points for a line, so we're now ready to graph. We essentially just have to connect the dots.
Graphing using intercepts (old)
So it's going to look something like this. It's going to look something-- our line. I don't have a good line tool, so I'm going to try my best to draw it nicely-- is going to look something like that. And you just keep going. You just keep going. The point-slope form is easiest to get, so let's do that first: Let's do both for clarity With algebra we have: The point slope form is obviously very easy to get when you have a point and the slope of the line which is the whole reason for this form.
The standard form takes a little more work but is very useful for drawing and analyzing your line. Hence the need for both. You can use either way you like to go between the two forms. It's all a matter of personal preference. We will do one more quickly and you should be able to answer the rest.
Graphing Calculator - MathPapa
Let's get the two forms of the line with point and slope given by: By direct substitution, we can get the point-slope form: This requires you to calculate the slope of the line from these points and then use the slope and one of the points as we did above to get the equation of a line.
So we need a formula for the slope given two points. There are several ways to get this. You can remember that the slope is the rise change in y divided by the run change in x or you could calculate this form from the point-slope form of a line.
If we have points x1, y1 and x2, y2by memory we have: Lets try an example: You gave the following two points: