How do you find the slope of the secant lines of #f(x)=(x^2)-4x# at [0,4]?

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Answer 1 · 2016-06-05T19:30:28

the secant line has slope #0#

Find the points at #x=0# and #x=4#.

#f(0)=0^2-4(0)=0" "=>" point at (0,0)"#

#f(4)=4^2-4(4)=0" "=>" point at (4,0)"#

The slope of the line connecting the points #(0,0)# and #(4,0)# is

#m=(0-0)/(4-0)=0/4=0#

The secant line is a horizontal line with slope #0#:

graph{(y-x^2+4x)(y-0)=0 [-9.375, 10.625, -5.12, 4.88]}

We can generalize the previous process to say that the slope of the secant line of #f(x)# at #x=a# and #x=b# is

#m=(f(b)-f(a))/(b-a)#