
Differential
calculus 


Differential
Calculus
 Derivatives and differentials 
Determining
the lines tangent to the graph of a function from a point outside the
function 
Determining
the lines tangent to the graph of a function from a point outside the
function, example 





Determining the lines tangent to the graph of a function from a point outside the
function

Lines
tangent to the graph of a function y
= f (x)
from a given point (x_{1},
y_{1})
outside the function are defined
by two
points they pass through, the given point
(x_{1},
y_{1})
and the point of tangency (x_{0},
y_{0}).

To
find the coordinates x_{0}
and y_{0}
of the points of tangency we should solve the following two
equations,

(1) y_{0}
= f (x_{0}) 
(2) y_{1}

y_{0}
= f
'
(x_{0}) · (x_{1}

x_{0})

as
the system of the two equations with the two unknowns x_{0}
and y_{0}.

The
first equation states that the point of tangency must lie on the
given function while, at the same time, its coordinates
x_{0}
and y_{0}
must satisfy the equation of the tangent line (2),
since the point of tangency is the common point of the
tangent
and the curve.

Thus
for example, if given is the quadratic function
f (x)
= ax^{2}
+ bx
+ c
and
the point P_{1}(x_{1},
y_{1})
t hen, applying
the above system of equations we can calculate
the coordinates of the points of tangency
P_{0
}(x_{0},
y_{0})
directly from the formulas,



Example: 
Find equations of the lines tangent
to the function f
(x)
= 
x^{2}
+ 2x
+ 4
drawn
from the point 

(3,
2).

Solution: To
find the coordinates x_{0}
and y_{0}
of the points of tangency we should solve the system of equations.
Substitute given quantities, the point (3,
2)
and f
(x)
= 
x^{2}
+ 2x
+ 4
into equations, 


Let calculate the slope
f
'
(x_{0})
of the tangent line, the equation
(2).


Then
substitute the slope and the y_{0}
into second equation to get the points of tangency, 

Therefore, the points of tangency
are (2,
4)
and (4, 4). 
To
write the equations of tangents t_{1}
and t_{2}
we calculate 
their
slopes m_{1}
and m_{2}
from 
f
'
(x_{0})
= 
2x_{0}
+ 2. 
Thus,
m_{1}
=
f
'
(2)
= 
2
· 2
+ 2
= 
2,
m_{1}
= 
2, 
and m_{2}
=
f
'
(4)
= 
2
· 4
+ 2
= 
6,
m_{2}
= 
6. 
The
tangent
t_{1}
through the point (2,
4)
and m_{1}
= 
2, and 
the
tangent
t_{2}
through the point (4, 4)
and m_{2}
= 
6, 
t_{1} ::
y
= 
2x
+ 8
and
t_{2} ::
y
= 
6x
+ 20. 



Check
the coordinates of the points of tangency by plugging the
corresponding parameters into the above
formulas.









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