Question

Find the coordinates of the points on the graph of `y=3x^2-2x` at which tangent line is parallel to the line `y=10x`?

Answer 1

`(2,8)`

Explanation:

Two lines are parallel if they share the same slope.

Any line parallel to `y=10x` will have the slope 10.

Now, we have: `y=3x^2-2x`

We try to find `dy/dx`

=>`d/dx(y)=d/dx(3x^2-2x)`

We use the power rule:

`d/dx(x^n)=nx^(n-1)` where `n` is a constant.

`dy/dx=3*2x^(2-1)-2*1x^(1-1)`

=>`dy/dx=6x-2`

Now, we set `dy/dx=10`

=>`10=6x-2`

=>`12=6x`

=>`2=x`

We plug this value in to get our `y`, or `f(x)`.

`3(2)^2-2(2)=y`

=>`12-4=y`

=>`8=y`

Therefore, the point is at `(2,8)`

Answer 2

`(2,8)`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"`

`y=3x^2-2x`

`rArrdy/dx=6x-2`

`y=10xtom=10`

`rArr6x-2=10rArrx=2`

`x=2toy=12-4=8`

`rArr"coordinates of point "=(2,8)`