Answer:
y = 16(x +2)
Explanation:
The equation of the tangent to a curve at a point can be written using the point-slope form of the equation for a line. The point is (x, f(x)), and the slope is the derivative, f'(x).
Point and slope
The point where you want the tangent line has y-value ...
y = 2(-2)³ -8(-2) = 2(-8)+2(8) = 0 . . . . . point (x, y) = (-2, 0)
The slope at the point where you want the tangent is ...
y' = 2·3x² -8 = 6(-2)² -8 = 24 -8 = 16 . . . . . slope m = 16
Equation of the tangent line
The point-slope equation of a line is ...
y -k = m(x -h) . . . . . line with slope m through point (h, k)
y -0 = 16(x -(-2)) . . . . . line with slope 16 through (-2, 0)
y = 16(x +2) . . . . . simplified
An equation of the tangent line is ...
y = 16(x +2)