Question

The point P(9, −2) lies on the curve y = 2 8 − x . (a) If Q is the point x, 2 8 − x , find the slope of the secant line PQ (correct to six decimal places) for the following values of x.
(i)
8.9
mPQ =
(ii)
8.99
mPQ =
(iii)
8.999
mPQ =
(iv)
8.9999
mPQ =
(v)
9.1
mPQ =
(vi)
9.01
mPQ =
(vii)
9.001
mPQ =
(viii)
9.0001
mPQ =
(b)
Using the results of part (a), guess the value of the slope of the tangent line to the curve at
P(9, −2).
m =
(c)
Using the slope from part (b), find an equation of the tangent line to the curve at
P(9, −2).

Answer 1

In this question, we use the slope equation and its further calculation can be defined as follows:

Slope equation $=\bold{\frac{y_2-y_1}{x_2-x_1}}$

$\to \bold{f(x)= y=\frac{2}{8-x}}\\\\\\\to \bold{P(9,-2)=(x_2,y_2)}\\\\\to \bold{q(x_1, f(x))}$

Calculating the Slope:

$=\bold{\frac{-2-(\frac{2}{8-x})}{9-x}}$

$=\bold{\frac{-16+2x-2}{(9-x)(8-x)}}\\\\=\bold{\frac{(2x-18)}{(9-x)(8-x)}}\\\\=\bold{\frac{-2(9-x)}{(9-x)(8-x)}}\\\\=\bold{\frac{-2}{(8-x)}}$

When

$x=8.9\\\\m= \bold{\frac{-2}{(8-8.9)}} = \bold{\frac{-2}{(0.9)}}=-2.222222222$

When

$x=8.99\\\\m= \bold{\frac{-2}{(8-8.99)}} = \bold{\frac{-2}{(0.99)}}=-2.020202020202$

When  

$x=8.999 \\\\m= \bold{\frac{-2}{(8-8.999)}} = \bold{\frac{-2}{(0.999)}}=-2.002002$

When

$x=9.1\\\\m= \bold{\frac{-2}{(8-9.1)}} = \bold{\frac{-2}{-1.1}}=1.818181$

When

$x=9.01\\\\m= \bold{\frac{-2}{(8-9.01)}} = \bold{\frac{-2}{-1.01}}=1.98019802$

When

$x=9.001\\\\m= \bold{\frac{-2}{(8-9.001)}} = \bold{\frac{-2}{-1.001}}=1.998002$

When

$x=9.0001\\\\m= \bold{\frac{-2}{(8-9.0001)}} = \bold{\frac{-2}{-1.0001}}=1.99980002$

For point b:

$P(9, -2)\\\\m = 2$

For point c:

Line equation:

$\to \bold{y= m(x- x_1)+y_1}\\\\\to \bold{y= 2(x-9)+(-2)}\\\\\to \bold{y= 2x-18-2}\\\\\to \bold{y= 2x-20}$

Learn more:

brainly.com/question/13219315