For the function f(x) = -12x^2 + 6x - 8 find the equation of the tangent line at x= -1

Let, f(x) = y = -12x^2+6x-8

Substitute x = -1 in f(x) we get y = -26

y = -12x^2+6x-8

Differentiate ‘y’ w.r.t ‘x’ we get

dy/dx = -24x+6

By putting value of ‘x’ we get

dy/dx = 30

Slope(m)=dy/dx = 30

Equation of tangent => (y - y1)=m(x - x1)

Here, x1= -1 and y1 = -26 and m = 30

So, (y - (-26)) =30(x - (-1))

y+26=30(x+1)

y+26=30x+30

30x - y + 4 =0 is the equation of tangent at x = -1