Question

Find the equation of the line tangent to the parabola `y=x^2 -3x + 2` and perpendicular to the line `5x - 2y - 10=0` ?

Answer 1

`y = -2/5x - 79/50`

Explanation:

Start by differentiating.

`y' = 2x - 3`

We know that the derivative represents the instantaneous rate of change of the function at any given point on the function's domain. We are given that our tangent needs to be perpendicular to `5x - 2y - 10 = 0`. Convert this to `y = mx + b` to find the slope:

`5x - 2y - 10=0`

`-2y = 10 - 5x`

`y = 5/2x - 5`

The perpendicular slope will be the negative reciprocal of this, or `-2/5`. You can solve for `x` now:

`-2/5 = 2x- 3`

`13/5 = 2x`

`x = 13/10`

So, the tangent will be perpendicular to the given line at that point.

Find the y-coordinate of this point:

`y = (13/10)^2 - 3(13/10) + 2 = 169/100 - 39/10 + 2 = -21/10`

Find the equation now:

`y - y_1 = m(x- x_1)`

`y - (-21/10) = -2/5(x - 13/10)`

`y + 21/10 = -2/5x + 26/50`

`y = -2/5x -79/50`

Hopefully this helps!