To find slope of the tangent line drawn at a particular point, we have to follow the procedure.
Find the slope of the tangent line at a point (x1, y1) by finding the derivative of the curve.
Apply the given point in the derivative to get the slope exactly a the point (x1, y1)
Using the formula
(y - y1) = m (x - x1)
We can find equation of the tangent line.
For each problem, find the equation of the tangent line
to the function at the given point.
Problem 1 :
f(x) = x2 + 1; (1, 2)
A) y = 2x C) y = (1/2)x + 3/2 |
B) y = -8x + 10 D) y = -6x + 8 |
Solution :
Given, f(x) = x2 + 1
f'(x) at (1, 2)
x = 1 and y = 2
f'(x) = m = 2x
m = 2(1)
Slope m = 2
Equation of a tangent line is y - y1 = m(x - x1)
x1 = 1 and y1 = 2
y - 2 = 2(x - 1)
y - 2 = 2x - 2
y = 2x - 2 + 2
y = 2x
Equation of a tangent line is y = 2x.
So, option A) is correct.
Problem 2 :
f(x) = 2x2 + 2x + 2; (-1, 2)
A) y = -8x – 6 C) y= (1/2)x + 5/2 |
B) y = -2x D) y = 6x + 8 |
Solution :
Given, f(x) = 2x2 + 2x + 2
f'(x) at (-1, 2)
x = -1 and y = 2
f'(x) = m = 4x + 2
m = 4(-1) + 2
m = -4 + 2
Slope m = -2
Equation of a tangent line is y - y1 = m(x - x1)
x1 = -1 and y1 = 2
y - 2 = -2(x + 1)
y - 2 = -2x - 2
y = -2x - 2 + 2
y = -2x
Equation of a tangent line is y = -2x.
So, option B) is correct.
Problem 3 :
y = x2 + x - 2; (1, 0)
C) y = 3x - 3 |
D) y = -12x + 12 |
Solution :
Given, y = x2 + x - 2
y' at (1, 0)
x = 1 and y = 0
y' = m = 2x + 1
m = 2(1) + 1
m = 2 + 1
Slope m = 3
Equation of a tangent line is y - y1 = m(x - x1)
x1 = 1 and y1 = 0
y - 0 = 3(x - 1)
y - 0 = 3x - 3
y = 3x - 3
Equation of a tangent line is y = 3x - 3.
So, option C) is correct.
Problem 4 :
y = 2x2 + 1; (-1, 3)
A) y = 2x + 5 B) y = -4x – 1 C) y = 12x + 15
Solution :
Given, y = 2x2 + 1
y' at (-1, 3)
x = -1 and y = 3
y' = m = 4x
m = 4(-1)
Slope m = -4
Equation of a tangent line is y - y1 = m(x - x1)
x1 = -1 and y1 = 3
y - 3 = -4(x + 1)
y - 3 = -4x - 4
y = -4x - 4 + 3
y = -4x - 1
Equation of a tangent line is y = -4x - 1.
So, option B) is correct.
Problem 5 :
A) y = (1/4)x + 1 C) y = -x + 1 |
B) -2x + 1 D) y = 1 |
Solution :
Equation of a tangent line is y - y1 = m(x - x1)
x1 = 0 and y1 = 1
y - 1 = -1(x - 0)
y - 1 = -1x
y = -x + 1
Equation of a tangent line is y = -x + 1.
So, option C) is correct.
Problem 6 :
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Solution :
Hence, option B) is correct.
Problem 7 :
A) y = -1 C) y = -(1/2)x - 2 |
B) y = x + 1 D) y = 4x + 7 |
Solution :
Equation of a tangent line is y - y1 = m(x - x1)
x1 = -2 and y1 = -1
y + 1 = 1(x + 2)
y = x + 2 - 1
y = x + 1
Equation of a tangent line is y = x + 1.
So, option B) is correct.
Problem 8 :
Solution :
So, option A) is correct.
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May 21, 24 08:51 AM
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