How do you find the equation of the secant line through the points where X has the given values: #f(x)=(x^2)-4x#; x=0, x=4?

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Answer 1 · 2015-09-20T23:43:36

See the explanation.

#y = f(x) = x^2-4x#

At #x=0#, we get #y = f(0) = (0)^2-4(0) = 0#

At #x=4#, we get #y = f(4) = (4)^2-4(4) = 0#

This gives us two points on the secant line:

#(0,0)# and #(4,0)#

The slope of the line through those points is #(0-0)/(4-0) = 0/4=0#

The equation of the line through #(0,0)# with slope #0# is #y=0#. (It is the horizontal axis).