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

To graph a line passing through a point (3, 0) and slope 4, we will find the equation of the line first.

Since, equation of a line passing through a point (h, k) and slope 'm' is given by,

y - k = m(x - h)

Therefore, equation of a line passing through (3, 0) and slope 4 will be,

y - 0 = 4(x - 3)

y = 4x - 12

For y-intercept put x = 0 in the equation,

y = 4(0) - 12

y = -12

For x-intercept put y = 0 in the equation,

0 = 4x - 12

x = 3

Therefore, line will intersect x-axis at (3, 0) and y-axis at (0, -12).

              By joining these points we can find the graph of the line passing through (3, 0) and having slope = 4.

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Hi there!

Recall that slope-intercept form is:

y = mx + b

Where m = slope

In this instance, we are given a slope of 4, therefore:

y = 4x + b

Substitute in the x and y coordinates of the point given:

0 = 4(3) + b

0 = 12 + b

Subtract 12 from both sides:

-12 = b

Therefore, the equation would be:

y = 4x - 12

Graph the equation by finding x and y values or using a calculator:

x = 0, y = 4(0) - 12 = 12            (0, 12)  

x = 1, y = 4(1) - 12 = -8               (1, -8)

x = 2, y = 4(2) - 12 = -4             (2, -4)

x = 3, y = 4(3) - 12 = 0              (3, 0)

And so forth:

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