Let \(f(x)=6 \sqrt{x+1}+5\).
Find the gradient of the tangent to \(y=f(x)\) at \(x=8\). (2 marks)
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Calculus, 2ADV C1 EQ-Bank 13
- Use differentiation by first principles to find \(y^{′}\), given \(y = 2x^2 + 5x\). (2 marks)
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- Find the equation of the tangent to the curve when \(x = 1\). (1 mark)
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Calculus, 2ADV C1 EQ-Bank 7 MC
At which point on the curve \(y=2x^{2}-11x+3\) can a tangent be drawn such that it is inclined at 45° when it crosses the positive \(x\)-axis?
- \((-3,54)\)
- \((-2,33)\)
- \((2,-11)\)
- \((3,-12)\)
Calculus, 2ADV C1 2023 HSC 14
Find the equation of the tangent to the curve `y=(2x+1)^3` at the point `(0,1)`. (3 marks)
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Calculus, 2ADV C1 2022 HSC 5 MC
Let `h(x)=(f(x))/(g(x))`, where
`{:[f(1)=2, qquad f^{′}(1)=4],[g(1)=8, qquad g^{′}(1)=12]:}`
What is the gradient of the tangent to the graph of `y=h(x)` at `x=1` ?
- `-8`
- `\ \ \ 8`
- `- 1/8`
- `\ \ \ 1/8`
Calculus, 2ADV C1 EQ-Bank 16
- Find the equations of the tangents to the curve `y = x^2-3x` at the points where the curve cuts the `x`-axis. (2 marks)
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- Where do the tangents intersect? (2 marks)
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Calculus, 2ADV C1 2017 HSC 12a
Find the equation of the tangent to the curve `y = x^2 + 4x - 7` at the point `(1, -2)`. (2 marks)
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Calculus, 2ADV C1 2009 HSC 1d
Find the gradient of the tangent to the curve `y = x^4- 3x` at the point `(1, –2)`. (2 marks)
Calculus, 2ADV C1 2010 HSC 7b
The parabola shown in the diagram is the graph `y = x^2`. The points `A (–1,1)` and `B (2, 4)` are on the parabola.
- Find the equation of the tangent to the parabola at `A`. (2 marks)
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- Let `M` be the midpoint of `AB`.
There is a point `C` on the parabola such that the tangent at `C` is parallel to `AB`.
Show that the line `MC` is vertical. (2 marks)
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- The tangent at `A` meets the line `MC` at `T`.
Show that the line `BT` is a tangent to the parabola. (2 marks)
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Show Worked Solution
| i. |  |
`y=x^2\ \ =>\ \ dy/dx=2x`
`text(At)\ \ A(-1,1)\ \ =>\ \ dy/dx = -2`
`text(T)text(angent has)\ \ m=text(–2),\ text(through)\ text{(–1,1):}`
| `y-y_1` | `= m(x-x_1)` |
| `y-1` | `= -2 (x + 1)` |
| `y-1` | `= -2x-2` |
| `2x + y + 1` | `= 0` |
`:.\ text(T)text(angent at)\ A\ text(is)\ \ 2x + y + 1 = 0`
♦ Mean mark 37%.
IMPORTANT: Key strategy for solution: the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).
IMPORTANT: Key strategy for solution: the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).
ii. `Atext{(–1,1)}\ \ \ B(2,4)`
`M= ((-1+2)/2 , (1+4)/2)= (1/2, 5/2)`
`m_(AB)= (y_2-y_1)/(x_2-x_1)= (4-1)/(2 + 1)=1`
`text(When)\ \ dy/dx=1:`
`2x=1\ \ =>\ \ x=1/2\ \ =>\ \ C \ (1/2, 1/4)`
`=>M\ text(and)\ C\ text(both have)\ x text(-value)=1/2`
`:. MC\ text(is vertical … as required)`
iii. `T\ text(is point on tangent when) \ x=1/2`
♦♦ Mean mark 29%.
`text(T)text(angent)\ \ \ 2x + y + 1 = 0`
`text(At)\ x = 1/2`
`2 xx (1/2) + y + 1=0\ \ =>\ \ y=–2`
`:.\ T (1/2, –2)`
`text (Given)\ B (2, 4):`
`m_(BT)= (4+2)/(2-1/2)=4`
`text(At)\ \ B(2,4),\ text(find gradient of tangent:)`
`dy/dx = 2x=2 xx2=4`
`m_text(tangent) = 4=m_(BT)`
`:.BT\ text(is a tangent)`
Calculus, 2ADV C1 2011 HSC 2c
Find the equation of the tangent to the curve `y = (2x + 1)^4` at the point where `x = -1`. (3 marks)
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Calculus, 2ADV C1 2012 HSC 11c
Find the equation of the tangent to the curve `y = x^2` at the point where `x = 3`. (2 marks)
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