Find the equation of the tangent to the curve $y = (2x + 1)^{3}$ at the point $(0, 1)$. - HSC - SSCE Mathematics Advanced - Question 14 - 2023 - Paper 1
Question 14
Find the equation of the tangent to the curve $y = (2x + 1)^{3}$ at the point $(0, 1)$.
Worked Solution & Example Answer:Find the equation of the tangent to the curve $y = (2x + 1)^{3}$ at the point $(0, 1)$. - HSC - SSCE Mathematics Advanced - Question 14 - 2023 - Paper 1
Step 1
Find the derivative of the function
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Answer
To find the equation of the tangent line, we first need to compute the derivative of the function ( y = (2x + 1)^{3} ). Using the chain rule:
[
y' = 3(2x + 1)^{2} \times 2
]
Thus, the derivative is:
[
y' = 6(2x + 1)^{2} ]
Step 2
Evaluate the derivative at the point (0, 1)
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Answer
Next, we evaluate the derivative at ( x = 0 ):
[
y' = 6(2(0) + 1)^{2} = 6(1)^{2} = 6
]
Therefore, the slope of the tangent at the point ( (0, 1) ) is ( 6 ).
Step 3
Use the point-slope form to find the tangent equation
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Answer
We can use point-slope form to find the equation of the tangent line:
[
y - y_1 = m(x - x_1)
]
where ( m = 6 ) is the slope, and ( (x_1, y_1) = (0, 1) ):
[
y - 1 = 6(x - 0)
]
This simplifies to:
[
y = 6x + 1
]
Thus, the equation of the tangent line is ( y = 6x + 1 ).
