# Secant Lines

Discussion in 'Calculus' started by nycmathguy, Oct 5, 2021.

1. ### nycmathguy

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Exercises 1.1
8 (a - c)

nycmathguy, Oct 5, 2021
2. ### MathLover1

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Consider the function f(x)=6x-x^2 and the point P(2,8) on the graph of f.

a.
graph f and secant lines passing through P(2,8) and Q(x, f(x)) for x-values 3,2.5, and 1.5.

The function is given by:

f(x)=6x-x^2

at x=3

f(3)=6*3-3^2=18-9=9 -> Q(3, 9)

at x =2.5
f(2.5)=6*2.5-2.5^2=8.75-> Q(2.5, 8.75)

at x =1.5
f(1.5)=6*1.5-1.5^2=6.75-> Q(1.5, 6.75)

find equations of the secant lines
using the slope-intercept form, m[1]=1, and Q(3, 9) we have
y-9= 1(x-3)
y= x-3+9
y= x+6-> first secant line

m[2]=2.5, and Q(2.5, 8.75) we have
y-8.75= 2.5(x-2.5)
y= 2.5x-6.25+8.75
y= 2.5x+2.5-> second secant line

m[3]=1.5, and Q(1.5, 6.75) we have
y-6.75= 1.5(x-1.5)
y= 1.5x-2.25+6.75
y= 1.5x+4.5-> third secant line

b.
Find the slope of each secant line

the slope of the secant line at x=3 is
m[1]=(9-8)/(3-2)
m[1]=1
the slope of the secant line at x=2.5 is Q(2.5, 8.75)
m[2]=(8.75-8)/(2.5-2)
m[2]=1.5

the slope of the secant line at x=1.5 is
m[3]=(6.75-8)/(1.5-2)
m[3]=2.5

c.
use the results of part b to estimate the slope of the tangent line to the graph of f at P(2,8) . Describe how to improve approximation of the slope.

Estimated slope of the tangent line at (2, 8) is 2 (between m[2] and m[3]). You can improve the approximation by decreasing the
distance between each point and (2, 8).

Last edited: Oct 5, 2021
MathLover1, Oct 5, 2021
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