The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − 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) 6.9
mPQ = 1
(ii) 6.99
mPQ = 2
(iii) 6.999
mPQ = 3
(iv) 6.9999
mPQ = 4
(v) 7.1
mPQ = 5
(vi) 7.01
mPQ = 6
(vii) 7.001
mPQ = 7
(viii) 7.000
mPQ = 8
(b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at
P(7, −2).
m = 9
(c) Using the slope from part (b), find an equation of the tangent line to the curve at
P(7, −2).

Respuesta :

Answer:

a) (i) [tex]m = 2.22[/tex], (ii) [tex]m = 2[/tex], (iii) [tex]m = 2[/tex], (iv) [tex]m = 2[/tex], (v) [tex]m = 1.82[/tex], (vi) [tex]m = 2[/tex], (vii) [tex]m = 2[/tex], (viii) [tex]m = 2[/tex]; b) [tex]m \approx 2[/tex]; c) The equation of the tangent line to curve at P (7, -2) is [tex]y = 2\cdot x + 12[/tex].

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

[tex]m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}[/tex]

(i) [tex]x_{Q} = 6.9[/tex]:

[tex]y_{Q} =\frac{2}{6-6.9}[/tex]

[tex]y_{Q} = -2.222[/tex]

[tex]m = \frac{-2.222 + 2}{6.9-7}[/tex]

[tex]m = 2.22[/tex]

(ii) [tex]x_{Q} = 6.99[/tex]

[tex]y_{Q} =\frac{2}{6-6.99}[/tex]

[tex]y_{Q} = -2.020[/tex]

[tex]m = \frac{-2.020 + 2}{6.99-7}[/tex]

[tex]m = 2[/tex]

(iii) [tex]x_{Q} = 6.999[/tex]

[tex]y_{Q} =\frac{2}{6-6.999}[/tex]

[tex]y_{Q} = -2.002[/tex]

[tex]m = \frac{-2.002 + 2}{6.999-7}[/tex]

[tex]m = 2[/tex]

(iv) [tex]x_{Q} = 6.9999[/tex]

[tex]y_{Q} =\frac{2}{6-6.9999}[/tex]

[tex]y_{Q} = -2.0002[/tex]

[tex]m = \frac{-2.0002 + 2}{6.9999-7}[/tex]

[tex]m = 2[/tex]

(v) [tex]x_{Q} = 7.1[/tex]

[tex]y_{Q} =\frac{2}{6-7.1}[/tex]

[tex]y_{Q} = -1.818[/tex]

[tex]m = \frac{-1.818 + 2}{7.1-7}[/tex]

[tex]m = 1.82[/tex]

(vi) [tex]x_{Q} = 7.01[/tex]

[tex]y_{Q} =\frac{2}{6-7.01}[/tex]

[tex]y_{Q} = -1.980[/tex]

[tex]m = \frac{-1.980 + 2}{7.01-7}[/tex]

[tex]m = 2[/tex]

(vii) [tex]x_{Q} = 7.001[/tex]

[tex]y_{Q} =\frac{2}{6-7.001}[/tex]

[tex]y_{Q} = -1.998[/tex]

[tex]m = \frac{-1.998 + 2}{7.001-7}[/tex]

[tex]m = 2[/tex]

(viii)  [tex]x_{Q} = 7.0001[/tex]

[tex]y_{Q} =\frac{2}{6-7.0001}[/tex]

[tex]y_{Q} = -1.9998[/tex]

[tex]m = \frac{-1.9998 + 2}{7.0001-7}[/tex]

[tex]m = 2[/tex]

b) The slope at P (7,-2) can be estimated by using the following average:

[tex]m \approx \frac{f(6.9999)+f(7.0001)}{2}[/tex]

[tex]m \approx \frac{2+2}{2}[/tex]

[tex]m \approx 2[/tex]

The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

[tex]y = m\cdot x + b[/tex]

Where:

[tex]x[/tex] - Independent variable.

[tex]y[/tex] - Depedent variable.

[tex]m[/tex] - Slope.

[tex]b[/tex] - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

[tex]-2 = 2 \cdot 7 + b[/tex]

[tex]b = -2 + 14[/tex]

[tex]b = 12[/tex]

The equation of the tangent line to curve at P (7, -2) is [tex]y = 2\cdot x + 12[/tex].