Question

Point A and B lie on the curve `y=x^2-4x+7`. Point A has coordinates `(4,7)` and B is the stationary point of the curve. The equation of a line L is y=mx-2 Where m is a constant. In the case where L passes through the midpoint of AB find the value of m?

Answer 1

`7/3`.

Explanation:

At the stationary point, `dy/dx=0`.

`:. d/dx{x^2-4x+7}=0`.

`:. 2x-4=0`.

`:. x=2`.

The corresponding `y=x^2-4x+7=2^2-4xx2+7=3`.

`:. B=B(2,3)`.

Given that, `A=A(4,7)`.

`:." The mid-point "M" of "AB" is, given by, "`

`M=M((2+4)/2,(3+7)/2)=M(3,5).`

`because, M in L : y=mx-2, :., 5=m(3)-2," giving,"`

`m=7/3`.

Enjoy Maths.!

Answer 2

`m = 7/3`

Explanation:

The function will have a stationary point when `dy/dx = 0`.

`dy/dx= 2x - 4`

So

`0 = 2x -4 -> 4 = 2x -> x = 2`

The corresponding value of `y` is`y = 2^2 - 4(2) + 7 = 3`

We now must find the midpoint between `(2, 3)` and `(4, 7)`.

`M = ((2 + 4)/2, (3 + 7)/2) = (3, 5)`

We can now effectively solve for `m`:

`5 = 3m - 2`

`m = 7/3`

Hopefully this helps!

Answer 3

`m = 7/3`

Explanation:

To find the stationary point, `B`, we must compute the first derivative:

`y' = 2x-4`

We can find the x-coordinate of the stationary point, `B`, by setting the first derivative equal to 0 and the solving for x:

`0 = 2x-4`

`2x=4`

`x = 2`

Find the corresponding y value by evaluating the function at `x = 2`:

`y = 2^2-4(2) + 7`

`y = 3`

Point `B = (2,3)`

The midpoint between `(4,7)` and `(2,3)` is:

`((4+2)/2,(7+3)/2) = (3,5)`

Substitute the point `(3,5)` into the line `y = mx-2` and solve for `m`:

`5=m(3)-2`

`7 = 3m`

`m = 7/3`