How do you graph #y = x^2 – 4x# by plotting points?

1 Answer
smendyka
Feb 5, 2018

See a solution process below:

Explanation:

First, let's determine a series of points to plot:

For x = 0:

#y = 0^2 - (4 * 0) = 0 - 0 = 0# the point is: #(0, 0)#

For x = 2:

#y = 2^2 - (4 * 2) = 4 - 8 = -4# the point is: #(2, -4)#

For x = -2:

#y = -2^2 - (4 * -2) = 4 + 8 = 12# the point is: #(-2, 12)#

For x = 4:

#y = 4^2 - (4 * 4) = 16 - 16 = 0# the point is: #(4, 0)#

For x = -4:

#y = 4^2 - (4 * -4) = 16 + 16 = 32# the point is: #(-4, 32)#

For x = 6:

#y = 6^2 - (4 * 6) = 36 - 24 = 12# the point is: #(6, 12)#

For x = 8:

#y = 8^2 - (4 * 8) = 64 - 32 = 32# the point is: #(8, 32)#

Now, let's plot the points:

graph{((x-6)^2 + (y-12)^2 - 1)((x-8)^2 + (y-32)^2 - 1)(x^2 + y^2 - 1)((x-2)^2 + (y+4)^2 - 1)((x+2)^2 + (y-12)^2 - 1)((x-4)^2 + y^2 - 1)((x+4)^2 + (y-32)^2 - 1) = 0 [-60, 60, -10, 50]}

We can now draw a line connecting the plotted points:

graph{(y - x^2 + 4x)((x-6)^2 + (y-12)^2 - 1)((x-8)^2 + (y-32)^2 - 1)(x^2 + y^2 - 1)((x-2)^2 + (y+4)^2 - 1)((x+2)^2 + (y-12)^2 - 1)((x-4)^2 + y^2 - 1)((x+4)^2 + (y-32)^2 - 1) = 0 [-60, 60, -10, 50]}

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