Question

Y=3(x); x=9; x=16

a. Find an equation for the secant line through the points where x has the given values. b. Find an equation for the line tangent to the curve when x has the first value.

Answer 1

a. y=3x

b. y=-x/3+30

Explanation:

Data

y=3x; x1=9 and x2=16

knowing that the secant line equation is expressed as y-y1 = m(x-x1) , we have:

y=f(x)=2x, y1=f(x1)=3*9=27 ; y2=f(x2)=3*16=48, so P1=(9,27) P2=(16,48)

m = (y2-y1)/(x2-x1) = (48-27)/(16-9) = 21/7 = 3; so

y-27 = 3(x-9) ⇒ y-27=3x-27 ⇒ y=3x

and for the tangent line equation the slope is -1/m, therefore when x1=9

y-y1=(-1/m)(x-x1) ⇒ y-27 = (-1/3)(x-9) ⇒ y-27 = (-x/3)+3 ⇒ y=-x/3+30