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

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- Jan 8th 2008, 06:38 PMchibiusagiintersections
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

- Jan 8th 2008, 06:46 PMJhevon
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 $\displaystyle (x_1,y_1) = (2,0)$ - Jan 8th 2008, 06:50 PMtopsquark
$\displaystyle y = (x - 2)(x^2 + 2x + 6)$

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