The formula given below can be used to find the equation of a tangent line to a curve.
(y - y_{1}) = m(x - x_{1})
Here m is the slope of the tangent line and (x_{1}, y_{1}) is the point on the curve at where the tangent line is drawn.
Example 1 :
Find the equation of the tangent line at x = 5 to the curve f(x) = √(2x - 1).
Solution :
f(x) = √(2x - 1)
Substitute x = 5.
f(5) = √9
= 3
y = 3
The point on the curve at where the tangent line is drawn is (5, 3).
Find the slope of the tangent line at ay x = 3 :
f(x) = √(2x - 1)
f'(x) = {(1/[2√(2x-1)]} ⋅ 2
f'(x) = 1/√(2x - 1)
Substitute x = 5.
f'(5) = 1/√9
slope (m) = 1/3
Equation of the tangent line :
y - y_{1} = m(x - x_{1})
Substitute m = 1/3 and (x_{1}, y_{1}) = (5, 3).
y - 3 = (1/3)(x - 5)
3y - 9 = x - 5
x - 3y - 5 + 9 = 0
x - 3y + 4 = 0
Example 2 :
If the line y = 3x + 9 is tangent to the graph of the function f at (2, 15), what is f'(2) ?
Solution :
Equation of the tangent line y = 3x + 9 is in slope-intercept form.
Comparing y = mx + b and y = 3x + 9, we get
slope (m) = 3
The first derivative of a function always represents the slope.
Since the tangent line is drawn at (2, 15),
slope at (2, 15) = 3
f'(2) = 3
Example 3 :
What is the x-coordinate of the point where the tangent line to the curve y = x^{2 }+ 12x + 11 is parallel to x-axis.
Solution :
y = x^{2 }+ 12x + 11
Find the first derivative to get the slope of the tangent line.
dy/dx = 2x + 12
Slope of any line which is parallel to x-axis is equal to zero.
Since the tangent line is parallel to x-axis, its slope is equal to zero.
dy/dx = 0
2x + 12 = 0
2x = -12
x = -6
Example 4 :
Find the equation of the tangent line which goes through the point (2, -1) and is parallel to the line given by the equation 2x - y = 1.
Solution :
2x - y = 1
Write the above equation in slope-intercept form :
-y = -2x + 1
y = 2x - 1
Comparing y = mx + b and y = 2x - 1, we get
slope (m) = 2
Since the tangent line is parallel to the line 2x - y = 1, the slopes are equal.
slope of the tangent line = 2
The tangent line is passing through the point (2, -1) with the slope 2.
Equation of the tangent line :
y - y_{1} = m(x - x_{1})
Substitute m = 2 and (x_{1}, y_{1}) = (2, -1).
y + 1 = 2(x - 2)
y + 1 = 2x - 4
2x - y - 5 = 0
Example 5 :
For a particular g, we know g'(5) = 2 and g(5) = 3. Write the equation of a tangent to g at x = 5.
Solution :
g'(5) = 2 and g(5) = 3
Slope of the tangent to g at x = 5 is 2.
We draw the tangent line at the point (5, 3).
Equation of the tangent line :
y - y_{1} = m(x - x_{1})
Substitute m = 2 and (x_{1}, y_{1}) = (5, 3).
y - 3 = 2(x - 5)
y - 3 = 2x - 10
2x - y - 7 = 0
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