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Graph a line with a slope of 4 that contains the point (3,0)

Sagot :

This Is how you would graph the line
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We are given the slope of the line and a point that it passes through.

So, we can use the Point-Slope Formula:

[tex]\rm{y-y1=m(x-x1)[/tex]

Where y1 is the y-coordinate of the point, m is the slope of the line, and x1 is the x-coordinate of the line.

Now, plug in the values:

[tex]\rm{y-0=4(x-3)[/tex]

[tex]\rm{y-0=4x-12[/tex]

[tex]\rm{y=4x-12[/tex] (the equation of the line in slope-intercept form)

Now, how to graph it?

First of all, the y-intercept is (0, -12)

Remember, y-intercepts always have an x-coordinate of 0.

Now, 4 is the slope (Rise/Run)

Where Rise is how many units we move up or down, and Run is how many units we move left or right.

In this case, we move 4 up, over 1, up 4, over 1, and so on, until we have a line.

Then all we have to do is take a ruler and connect the points. :)

Hope you find it helpful.

Feel free to ask if you have any doubts.

[tex]\bf{-Sparkles^**^*^[/tex]

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