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
we can find how many zeros the function has by simply solving for them.
set $\displaystyle (x - 2) \left( x^2 + 2x + 6 \right) = 0$
$\displaystyle \Rightarrow x - 2 = 0$ or $\displaystyle x^2 + 2x + 6 = 0$
$\displaystyle \Rightarrow x = 2$ or $\displaystyle x = \frac {-2 \pm \sqrt{-20}}2$
clearly the quadratic has no real roots, thus the only root is $\displaystyle x = 2$
to find the tangent line, use the point-slope form.
$\displaystyle y - y_1 = m(x - x_1)$
here, $\displaystyle m = f'(2)$ and $\displaystyle (x_1,y_1) = (2,0)$
$\displaystyle y = (x - 2)(x^2 + 2x + 6)$
When this crosses the x-axis we have y = 0. So to find these x values:
$\displaystyle (x - 2)(x^2 + 2x + 6) = 0$
So either
$\displaystyle x - 2 = 0 \implies x = 2$
or
$\displaystyle x^2 + 2x + 6 = 0$
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.
$\displaystyle y^{\prime}(x) = 3x^2 + 2$
So at x = 2 the slope of the tangent to the curve is $\displaystyle y^{\prime}(2) = 14$.
So we need the equation of a line with a slope of 14 that passes through the point (2, 0).
$\displaystyle y = 14x + b$
$\displaystyle 0 = 14 \cdot 2 + b \implies b = -28$
So the tangent line at (2, 0) is $\displaystyle y = 14x - 28$.
-Dan