Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point
When this crosses the x-axis we have y = 0. So to find these x values:
Note, however, that the quadratic factor here has no real zeros. So no real zeros are obtained by solving this equation.
Thus the curve only crosses the x-axis once.
Now to find the equation of the tangent.
So at x = 2 the slope of the tangent to the curve is .
So we need the equation of a line with a slope of 14 that passes through the point (2, 0).
So the tangent line at (2, 0) is .