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

Step-by-step 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.

$\frac{d}{dx} ( {x}^{2} - x - 12) = 2x - 1$

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

$y = - 7x - 21$

$y = 7x - 28$