Question

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).

Answer 1

Answer:

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

Step-by-step explanation:

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

$m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$

(i) $x_{Q} = 6.9$:

$y_{Q} =\frac{2}{6-6.9}$

$y_{Q} = -2.222$

$m = \frac{-2.222 + 2}{6.9-7}$

$m = 2.22$

(ii) $x_{Q} = 6.99$

$y_{Q} =\frac{2}{6-6.99}$

$y_{Q} = -2.020$

$m = \frac{-2.020 + 2}{6.99-7}$

$m = 2$

(iii) $x_{Q} = 6.999$

$y_{Q} =\frac{2}{6-6.999}$

$y_{Q} = -2.002$

$m = \frac{-2.002 + 2}{6.999-7}$

$m = 2$

(iv) $x_{Q} = 6.9999$

$y_{Q} =\frac{2}{6-6.9999}$

$y_{Q} = -2.0002$

$m = \frac{-2.0002 + 2}{6.9999-7}$

$m = 2$

(v) $x_{Q} = 7.1$

$y_{Q} =\frac{2}{6-7.1}$

$y_{Q} = -1.818$

$m = \frac{-1.818 + 2}{7.1-7}$

$m = 1.82$

(vi) $x_{Q} = 7.01$

$y_{Q} =\frac{2}{6-7.01}$

$y_{Q} = -1.980$

$m = \frac{-1.980 + 2}{7.01-7}$

$m = 2$

(vii) $x_{Q} = 7.001$

$y_{Q} =\frac{2}{6-7.001}$

$y_{Q} = -1.998$

$m = \frac{-1.998 + 2}{7.001-7}$

$m = 2$

(viii)  $x_{Q} = 7.0001$

$y_{Q} =\frac{2}{6-7.0001}$

$y_{Q} = -1.9998$

$m = \frac{-1.9998 + 2}{7.0001-7}$

$m = 2$

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

$m \approx \frac{f(6.9999)+f(7.0001)}{2}$

$m \approx \frac{2+2}{2}$

$m \approx 2$

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:

$y = m\cdot x + b$

Where:

$x$ - Independent variable.

$y$ - Depedent variable.

$m$ - Slope.

$b$ - 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:

$-2 = 2 \cdot 7 + b$

$b = -2 + 14$

$b = 12$

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