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

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Answer 1 · 2015-10-23T20:20:11

See the explanation section, below.

The secant line to the graph of #f(x) = -x^2#; through the points where #x=-2#, and #x=-4# is the line through the points:

#(-2,f(-2))# and #(-4, f(-4))#

Which are the points: #(-2,-4)# and #(-4,-16)#

The slope is #m = ((-16)-(-4))/((-4)-(-2)) = (-12)/-2 =6#

The line has equation #y+4=6(x+2)#

Or #y+16=6(x+4)#

Or #y=6x+8#