smc-973-10-Find Tangent Equation

Calculus, 2ADV C1 EQ-Bank 3 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?

\((-3,54)\)
\((-2,33)\)
\((2,-11)\)
\((3,-12)\)

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\(D\)

Show Worked Solution

\(y\)	\(=2x^{2}-11x+3\)
\(y^{′}\)	\(=4x-11\)

\(\text{If a tangent crosses (positive) x-axis at 45}^{\circ},\)

\(m_{\text{tang}}=1 \ \ (\tan 45^{\circ}=1) \)

\(\text{Find}\ x\ \text{when}\ \ y^{′}=1: \)

\(4x-11\)	\(=1\)
\(4x\)	\(=12\)
\(x\)	\(=3\)

\(\text{Tangent at point}\ (3,-12) \)

\(\Rightarrow D\)

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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`y=6x+1`

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`y`	`=(2x+1)^3`
`dy/dx`	`=3xx2(2x+1)^2`
	`=6(2x+1)^2`

`text{At}\ x=0\ \ =>\ \ dy/dx=6xx1^2=6`

`text{Find equation of line}\ \ m=6,\ text{through}\ (0,1)`

`y-y_1`	`=m(x-x_1)`
`y-1`	`=6(x-0)`
`y`	`=6x+1`

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`
`qquad 8`
`- 1/8`
`qquad 1/8`

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`D`

Show Worked Solution

`text{Using the quotient rule:}`

`h′(1)`	`=(g(1)\ f′(1)-f(1)\ g′(1))/[g(1)]^2`
	`=(8xx4-2xx12)/(8^2)`
	`=(32-24)/64`
	`=1/8`

`=>D`

Calculus, 2ADV C1 SM-Bank 2

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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`y = −3x`
`y = 3x – 9`
`(3/2, −9/4)`

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i. `y`	`= x^2 – 3x`
	`= x(x – 3)`

`text(Cuts)\ xtext(-axis at)\ \ x = 0\ \ text(or)\ \ x = 3`

`(dy)/(dx) = 2x – 3`

`text(At)\ \ x = 0 \ => \ (dy)/(dx) = -3`

`T_1\ text(has)\ \ m = −3,\ text{through (0, 0)}`

`y – 0`	`= -3(x – 0)`
`y`	`= -3x`

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 3`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y – 0`	`= 3(x – 3)`
`y`	`= 3x – 9`

ii. `text(Intersection occurs when:)`

`3x – 9`	`= -3x`
`6x`	`= 9`
`x`	`= 3/2`

`y = -3 xx 3/2 = −9/2`

`:.\ text(Intersection at)\ \ (3/2, −9/2)`

Calculus, 2ADV C1 EQ-Bank 3

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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`y^{′} = 4x + 5`
`y = 9x-2`

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i.	`f(x)`	`= 2x^2 + 5x`
	`f^{′}(x)`	`= lim_(h->0) (f(x + h)-f(x))/h`
		`= lim_(h->0) ((2(x + h)^2 + 5(x + h))-(2x^2 + 5x))/h`
		`= lim_(h->0)(2x^2 + 4xh + 2h^2 + 5x + 5h-2x^2-5x)/h`
		`= lim_(h->0)(4xh + 2h^2 + 5h)/h`
		`= lim_(h->0)(h(4x + 5 + 2h))/h`

`:.\ y^{′} = 4x + 5`

ii. `text(When)\ \ x = 1, y = 7`

`y^{′} = 4 + 5 = 9`

`:. y-7`	`= 9(x-1)`
`y`	`= 9x-2`

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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`y = 6x – 8`

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`y`	`= x^2 + 4x – 7`
`(dy)/(dx)`	`= 2x + 4`

`text(When)\ x = 1,\ \ (dy)/(dx) = 6`

`text(Equation of tangent through)\ (1, -2)`

`y + 2`	`= 6 (x – 1)`
`y`	`= 6x – 8`

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)

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`text(Gradient = 1.`

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`y`	`= x^4\ – 3x`
`dy/dx`	`= 4x^3\ – 3`

`text(At)\ x = 1`

`dy/dx = 4\ – 3 = 1`

`:.\ text(Gradient of tangent at)\ (1,–2) = 1.`

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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`2x + y + 1 = 0`
`text(Proof) text{(See Worked Solutions)}`
`text(Proof) text{(See Worked Solutions)}`

Show Worked Solution

`y`	`=x^2`
`dy/dx`	`= 2x`

`text(At)\ \ A text{(–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: The critical understanding required for this question is that 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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`8x + y + 7 = 0`

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`y = (2x + 1)^4`

`text{Using the Function of a Function Rule (or Chain Rule)}`

`dy/dx`	`= 4 xx (2x + 1)^3 xx d/dx (2x + 1)`
	`= 8 (2x + 1)^3`

`text(At)\ \ x = –1,\ y = 1`

MARKER’S COMMENT: The best setting out clearly showed the derivative function, the gradient, the point and then finally, calculations for the equation of the tangent, as per the Worked Solution.

`dy/dx`	`= 8 (2(–1) + 1)^3`
	`= 8 (–1)^3`
	`= -8`

`text(T)text(angent has)\ m = –8\ text(through)\ (–1,1)`

`text(Using)\ \ \ y – y_1`	`= m (x – x_1)`
`y -1`	`= -8 (x + 1)`
`y – 1`	`= -8x -8`
`8x + y + 7`	`= 0`

`:.\ text(Equation of tangent is)\ 8x + y + 7 = 0`

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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`6x-y-9 = 0 `

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`y`	`=x^2`
`dy/dx`	`=2x`

`text{Need to find equation with m = 6, through (3,9) }`

`text(When) \ x = 3, y = 9, dy/dx = 6`

`text(Using)\ \ \ y-y_1`	`= m (x-x_1)`
`y -9`	`= 6(x -3)`
`y -9`	`= 6x -18`
`6x -y -9`	`=0`

`:.\ text( Equation of the tangent is 6x-y-9 = 0)`