Respuesta :
Answer:
Find the slope of the line that passes through the points shown in the table.
The slope of the line that passes through the points in the table is
.
Step-by-step explanation:
This answer can be solved by using formula for finding slope.
Calculate the required answers:
Given that point P(6,-2) lies on the curve y = 2/(5 − x) and Q is a point
(x, 2/(5 − x))
(a) Slope formula = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex]
(1) x=5.9, y = 2/(5 − (5.9))= - 2.22222222
Q=(5.9,-2.22222222), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-2.22222222)-(-2) }{5.9 -6 }[/tex] = 0.2222/0.1 =2.22222222
(2) x=5.99, y = 2/(5 − (5.99))= - 2.02020202
Q=(5.99,-2.02020202), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-2.02020202)-(-2) }{5.99 -6 }[/tex] = 0.02020202/0.01 =2.020202
(3) x=5.999, y = 2/(5 − (5.999))= - 2.002002002
Q=(5.999,-2.2222...), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-2.002002002)-(-2) }{5.999 -6 }[/tex]=0.002002002/0.001 =2.002002
(4) x=5.9999, y = 2/(5 − (5.9999))= - 2.00020002
Q=(5.9999,-2.00020002), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-2.00020002)-(-2) }{5.9999 -6 }[/tex]=0.00020002/0.0001 =2.0002
(5) x=6.1, y = 2/(5 − (6.1))= - 1.818181818
Q=(6.1,-1.818181818), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-1.818181818)-(-2) }{6.1 -6 }[/tex]=0.1818181/0.1 =1.818181818
(6) x=6.01, y = 2/(5 − (6.01))= - 1.98019802
Q=(6.01,-1.98019802), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-1.98019802)-(-2) }{6.01 -6 }[/tex]=0.01980198/0.01 = 1.98019802
(7) x=6.001, y = 2/(5 − (6.001))= - 1.998001998
Q=(6.001,-1.998001998), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-1.998001998)-(-2) }{6.001-6 }[/tex]=0.001998001998/0.001 = 1.998001998
(8) x=6.0001, y = 2/(5 − (6.0001))= - 1.99980002
Q=(6.0001,-1.99980002), P(6,-2)
[tex]m_{PQ}[/tex] = [tex]\frac{y_{2}-y_{1} }{x_{2} -x_{1} }[/tex] = [tex]\frac{(-1.99980002)-(-2) }{6.0001 -6 }[/tex]=0.000199980002/0.0001 = 1.99980002
(b) From the results of part (a), we can say that m is negative of y₂
So, the slope m of the tangent line to the curve at P(6, −2) will be negative of (-2) that is 2.
(c) Equation of the tangent line to the curve at (x₁,y₁), y-y₁=m(x-x₁)
Equation of the tangent line to the curve at P(6, −2) is
y-(-2) = 2(x-6)
y+2 = 2x -12
y = 2x - 14
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