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).
+36916 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?
