Question

An equation of the secant line containing (4, h(4)) and (6, h(6)) is

Answer 1

y= (h(6)-h(4))/2x +C

Where C=-2*h(6)+3*h(4)

Explanation:

The secant line is the line that meets a function (in this case h(x) ) in two points, so we have to apply the ecuation of a straigt line that meets two points:

y-y1 = (y2-y1)/(x2-x1) * (x-x1)

In this case X1=4 , x2=6, y1 = h(4) and y2= h(6)

So

y-h(4)= 1/2 (h(6)-h(4)) * (x-4)

y-h(4)= 1/2 (h(6)-h(4)) x- 2 *(h(6)-h(4))

y-h(4)= 1/2 (h(6)-h(4)) x- 2 (h(6) + 2h(4))

y= 1/2 (h(6)-h(4)) x- 2 h(6) + 2h(4) + h(4)

y= 1/2 (h(6)-h(4)) x- 2 h(6) + 3h(4)

Good Luck!

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