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Calculus, 2ADV C1 EO-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2-5x+6`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

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  1. `y = −x+2`
    `y = x-3`
  2. `(5/2, −1/2)`
Show Worked Solution
a.   `y` `= x^2-5x+6`
  `= (x-2)(x-3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x-5`

 
`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0` `= -1(x-2)`
`y` `= -x+2`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

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

`y -0` `= 1(x-3)`
`y` `= x -3`

 

b.   `text(Intersection occurs when:)`

`-x+2` `= x-3`
`2x` `= 5`
`x` `= 5/2`

  

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 EO-Bank 1

Find the equation of the tangent to the curve  \(y=e^{x^2+3x}\)  at the point where \(x=1\).  (2 marks)

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`y = 5e^4x-4e^4`

Show Worked Solution
\(y\) \(=e^{x^2+3x}\)
`(dy)/(dx)` \(=(2x+3)e^{x^2+3x}\)

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 5e^4`

`text(Equation of tangent through)\ (1, e^4)`

`y-e^4` `= 5e^4(x – 1)`
`y` `= 5e^4x-4e^4`

Calculus, 2ADV C1 2023 HSC 14 v1

Find the equation of the tangent to the curve  `y=x(3x+2)^2`  at the point `(1,25)`.   (3 marks)

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`y=55x-30`

Show Worked Solution
`y` `=x(3x+2)^2`  
`dy/dx` `=6x(3x+2) + (3x+2)^2`  
  `=(3x+2)(9x+2)`  

 
`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`
 

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1` `=m(x-x_1)`  
`y-25` `=55(x-1)`  
`y` `=55x-30`  

Calculus, 2ADV C1 EO-Bank 3 MC

At which point on the curve  \(y = x^{2}-6x + 8\)  can a normal be drawn such that it is inclined at 45\(^{\circ}\) to the positive \(x\)-axis?

  1. \((1,3)\)
  2. \((2,0)\)
  3. \(\left(\dfrac{5}{2}, -\dfrac{3}{4}\right)\)
  4. \((5,-7)\)
Show Answers Only

\(C\)

Show Worked Solution

\(y = x^{2}-6x + 8\)

\(y^{′} = 2x-6\)

If the normal is inclined at \(45^{\circ}\) to the positive x-axis then:

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

\(\text{Since } m_{\text{tangent}} \times m_{\text{normal}} = -1,\)

\(\therefore m_{\text{tangent}} = -1.\)
 

Find \(x\) when \(y^{′} = -1:\)

\(2x-6\) \(=-1\)  
\(2x\) \(=5\)  
\(x\) \(=\dfrac{5}{2}\)  

 
Find \(y:\)

\(y\) \(= \left(\dfrac{5}{2}\right)^{2}-6\left(\dfrac{5}{2}\right) + 8\)  
  \(=\dfrac{25}{4}-15 + 8\)  
  \(= -\dfrac{3}{4}\)  

 
\(\Rightarrow C\)

Calculus, 2ADV C1 EQ-Bank 3

  1.  Use differentiation by first principles to find \(y^{′}\), given  \(y = 2x^2 + 5x\).   (2 marks)

  2.  Find the equation of the tangent to the curve when  \(x = 1\).   (1 mark)

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  1.  `y^{′} = 4x+5`
  2.  `y = 9x-2`
Show Worked Solution
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 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?

  1. \((-3,54)\)
  2. \((-2,33)\)
  3. \((2,-11)\)
  4. \((3,-12)\)
Show Answers Only

\(D\)

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`

Show Worked Solution
`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` ?

  1. `-8`
  2. `\ \ \ 8`
  3. `- 1/8`
  4. `\ \ \ 1/8`
Show Answers Only

`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

  1.  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)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −3x`
    `y = 3x – 9`
  2. `(3/2, −9/4)`
Show Worked Solution
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 EO-Bank 3

  1.  Use differentiation by first principles to find \(y^{′}\), given  \(y = 4x^2 - 5x + 4\).   (2 marks)

  2.  Find the equation of the tangent to the curve when  \(x = 3\).   (1 mark)

Show Answers Only
  1.  `y^{′} = 8x-5`
  2.  `y = 19x-32`
Show Worked Solution
i.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((4(x + h)^2-5(x + h) + 4)-(4x^2-5x + 4))/h`
    `= lim_(h->0)(4x^2 + 8xh + 4h^2-5x-5h + 4-4x^2+5x-4)/h`
    `= lim_(h->0)(8xh + 4h^2-5h)/h`
    `= lim_(h->0)(h(8x-5 + 4h))/h`

 
`:.\ y^{′} = 8x-5`
 

ii.   `text(When)\ \ x = 3, y = 25`

`y^{′} = 24-5 = 19`
 

`:. y-25` `= 19(x-3)`
`y` `= 19x-32`

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`

Show Worked Solution
`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 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.
 

 
 

  1.  Find the equation of the tangent to the parabola at `A`.   (2 marks)

  2.  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)  

  3. The tangent at `A` meets the line `MC` at `T`.

     

    Show that the line `BT` is a tangent to the parabola.  (2 marks)

Show Answers Only
  1. `2x + y + 1 = 0`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
Show Worked Solution
i.
`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`

Show Worked Solution

`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 `

Show Worked Solution
`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)`