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Help, I'm confused.

 

The point P(4, −2) lies on the curve y = 2/(3 − x).

(a) If Q is the point (x, 2/(3 − 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.

(i) 3.9 mPQ =

(ii) 3.99 mPQ =

(iii) 3.999 mPQ =

(iv) 3.9999 mPQ =

(v) 4.1 mPQ =

(vi) 4.01 mPQ =

(vii) 4.001 mPQ =

(viii) 4.0001 mPQ =

 

(b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(4, −2). m =

 

(c) Using the slope from part (b), find an equation of the tangent line to the curve at P(4, −2).

 Sep 2, 2020
 #1
avatar+27001 
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Remember slope =   (y1-y2)/(x1-x2)

2/(3-x)-(-2)/ (x-4) = slope

(2 + 2(3-x))/(3-x)   / (x-4)

(2 +6-2x) /[(3-x) (x-4)]

 (8-2x) / [(3-x)(x-4)]

  -2(x-4) / [(3-x)(x-4)]

   slope = -2 / (3-x)             can you take it from here?     

 Sep 3, 2020

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