Question

Find the equations of the tangents to the curve y = x²-x-12 at each of the points where the curve crosses the x-axis.​

Answer 1

Explanation:

First, find the zeroes of the parabola

`{x}^(2) - x - 12`

`{x}^(2) - 4x + 3x - 12`

`x(x - 4) + 3(x - 4)`

`(x + 3)(x - 4) = 0`

`x = - 3`

`x - 4 = 0`

`x = 4`

So the zeroes or where the curve crosses the x axis is at 4 and -3.

Now, we take the derivative of the function.

`(d)/(dx) ( {x}^(2) - x - 12) = 2x - 1`

Plug in -3, and 4 into the derivative function

`2( - 3) = - 7`

`2(4) - 1 = 7`

So at x=-3, our slope of the tangent line is -7 and must pass through (-3,0). So we use point slope formula.

`y - y _(1) = m(x - x _(1))`

`y - 0 = - 7(x - ( - 3)`

`y = - 7x - 21`

At x=4, our slope of tangent line is 7, and pass through (4,0) so

`y - 0 = 7(x - 4)`

`y = 7x - 28`

So the equations of tangent is

`y = - 7x - 21`

`y = 7x - 28`