The point P (3, 2) lies on the curve y = 2/(2 x ). (a) If Q is the point ( x , 2

Question

The point

P(3, 2)

lies on the curve

y = 2/(2 x).

(a) If Q is the point

(x, 2/(2 x)),

use your calculator to find the slope

mPQ

of the secant line PQ (correct to six decimal places) for the following values of x.

The point ,-2) lies on the curve y = 2/(2-x). (a) If Q is the point x, 2/(2 - x)), use your calculaor tofind the slope mpo of the secant line PQ (correct to six decimal places) for the following values of x mp = 2.469139 (ii) 2.99 mR2 = 2.020202 (iii) 2.999 MpQ = 1.400080 (iv) 2.9999 mPQ =-2.0002 (v) 3.1 mPQ·-1.818182 (vi) 3.0:1 mPQ = |-1.980198 (vii) 3.001 mp1.998002 (vii) 3.0001 mPQ = 1.9998 (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(3, -2) m= c) Using the slope from part (b), find an equation of the tangent line to the curve at P(3, -2

Explanation / Answer

a) We have given point P(3,-2) and Q(x,2/(2-x))

i) we have given x=2.9

the slope of secant line between two points is m=(y2-y1)/(x2-x1)

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(2.9,2/(2-2.9))=Q(2.9,-2/0.9)

mPQ=(y2-y1)/(x2-x1)=((-2/0.9)+2)/(2.9-3)=2.222222

ii) we have given x=2.99

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(2.99,2/(2-2.99))=Q(2.99,-2/0.99)

mPQ=(y2-y1)/(x2-x1)=((-2/0.99)+2)/(2.99-3)=2.020202

iii) we have given x=2.999

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(2.999,2/(2-2.999))=Q(2.999,-2/0.999)

mPQ=(y2-y1)/(x2-x1)=((-2/0.999)+2)/(2.999-3)=2.002002

iv) we have given x=2.9999

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(2.9999,2/(2-2.9999))=Q(2.9999,-2/0.9999)

mPQ=(y2-y1)/(x2-x1)=((-2/0.9999)+2)/(2.9999-3)=2.000200

v) we have given x=3.1

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(3.1,2/(2-3.1))=Q(3.1,-2/1.1)

mPQ=(y2-y1)/(x2-x1)=((-2/1.1)+2)/(3.1-3)=1.818181

vi) we have given x=3.01

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(3.01,2/(2-3.01))=Q(3.01,-2/1.01)

mPQ=(y2-y1)/(x2-x1)=((-2/1.01)+2)/(3.01-3)=1.980198

vii) we have given x=3.001

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(3.001,2/(2-3.001))=Q(3.001,-2/1.001)

mPQ=(y2-y1)/(x2-x1)=((-2/1.001)+2)/(3.001-3)=1.998001

viii) we have given x=3.0001

P(x1,y1)=P(3,-2) and Q(x2,y2)=Q(3.0001,2/(2-3.0001))=Q(3.0001,-2/1.0001)

mPQ=(y2-y1)/(x2-x1)=((-2/1.0001)+2)/(3.0001-3)=1.999800

b) you can guess the slope m=2 of the tangent line to the curve at P(3,-2)

c) We have slope m=2 and point P(x1,y1)=P(3,-2)

An equation of the tangent line to the curve at P(3,-2) is y-y1=m(x-x1)

y+2=2(x-3)

y=2x-6-2

y=2x-8

An equation of the tangent line to the curve at P(3,-2) is y=2x-8

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