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

Answer 1

Avery Alexander

See a solution process below:

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]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Genesis Allen

To graph $y = x^2 - 4x$ by plotting points, you can choose values for $x$ , plug them into the equation to find the corresponding $y$ values, and then plot those points on a coordinate plane. Repeat this process for multiple values of $x$ to get a clear picture of the curve.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.