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Calculus, 2ADV C1 EQ-Bank 16

  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

a.    `y = -3x, \ y = 3x-9`

b.    `(3/2, -9/4)`

Show Worked Solution

a.    `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`

 

b.   `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(-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`).

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)`