intersections

**chibiusagi** · January 8th 2008, 06:38 PM

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

**Jhevon** · January 8th 2008, 06:46 PM

Originally Posted by chibiusagi

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 $(x - 2) \left( x^2 + 2x + 6 \right) = 0$

$\Rightarrow x - 2 = 0$ or $x^2 + 2x + 6 = 0$

$\Rightarrow x = 2$ or $x = \frac {-2 \pm \sqrt{-20}}2$

clearly the quadratic has no real roots, thus the only root is $x = 2$

to find the tangent line, use the point-slope form.

$y - y_1 = m(x - x_1)$

here, $m = f'(2)$ and $(x_1,y_1) = (2,0)$

**topsquark** · January 8th 2008, 06:50 PM

Originally Posted by chibiusagi

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

$y = (x - 2)(x^2 + 2x + 6)$

When this crosses the x-axis we have y = 0. So to find these x values:
$(x - 2)(x^2 + 2x + 6) = 0$

So either
$x - 2 = 0 \implies x = 2$
or
$x^2 + 2x + 6 = 0$

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.

$y^{\prime}(x) = 3x^2 + 2$

So at x = 2 the slope of the tangent to the curve is $y^{\prime}(2) = 14$ .

So we need the equation of a line with a slope of 14 that passes through the point (2, 0).

$y = 14x + b$

$0 = 14 \cdot 2 + b \implies b = -28$

So the tangent line at (2, 0) is $y = 14x - 28$ .

-Dan

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