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secondary 4 | A Maths
2 Answers Below
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Date Posted: 4 years ago
Views: 532
y = 2e^3x
When x = 1, y = 2e^3
Now, dy/dx
= 6e^3x
= 6e^3 when x = 1
Gradient of tangent is 6e^3
Gradient of normal is -1/(6e^3)
For the tangent,
y = 6e^3 * x + c
2e^3 = 6e^3 * 1 + c
c = -4e^3
Tangent: y = 6e^3 * x - 4e^3
For the normal,
y = -1/(6e^3) * x + c
2e^3 = -1/(6e^3) * 1 + c
c = 2e^3 + 1/(6e^3)
Normal:
y = -1/(6e^3) * x + 2e^3 + 1/(6e^3)
(6e^3)y = -x + 12 (e^3)^2 + 1
(6e^3)y = -x + 12e^6 + 1
When x = 1, y = 2e^3
Now, dy/dx
= 6e^3x
= 6e^3 when x = 1
Gradient of tangent is 6e^3
Gradient of normal is -1/(6e^3)
For the tangent,
y = 6e^3 * x + c
2e^3 = 6e^3 * 1 + c
c = -4e^3
Tangent: y = 6e^3 * x - 4e^3
For the normal,
y = -1/(6e^3) * x + c
2e^3 = -1/(6e^3) * 1 + c
c = 2e^3 + 1/(6e^3)
Normal:
y = -1/(6e^3) * x + 2e^3 + 1/(6e^3)
(6e^3)y = -x + 12 (e^3)^2 + 1
(6e^3)y = -x + 12e^6 + 1
See 2 Answers
done
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clear
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Answer is as above.Please upvote for me!!!
Date Posted:
4 years ago
Where is the equation of the tangent?
done
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Equation of the tangent is shown as above.
Date Posted:
4 years ago
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