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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)

Show Answers Only
  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 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)

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  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 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)`