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3 years ago

find zeros and y intercept and describe the end behavior in proper notation. then sketch the graph. 3x^3+2x^2-12x-8

Mathematics

1 answer:

Evgen [1.6K]3 years ago

5 0

Answer: The zeros of the given equation are 2, -2 and -0.667. The y-intercept of the graph is f(0) = -8.

Explanation:

The given expression is,

$f(x)=3x^3+2x^2-12x-8$

To find zeros put f(x)=0

$3x^3+2x^2-12x-8=0$

This equation is satisfied by x=2, so (x-2) is the factor of the given equation. By synthetic division or long division we can find the remaining factor.

$(x-2)(3x^2+8x+4)=0$

Use Factoring method.

$(x-2)(3x^2+6x+2x+4)=0$

$(x-2)(3x(x+2)+2(x+2)=0$

$(x-2)(3x+2)(x+2)=0$

Equate each factor equal to zero by using zero product property.

So the zeros of the given equation are 2, -2 and -0.667.

To find the y-intercept of the graph, put x=0.

$f(0)=3(0)^3+2(0)^2-12(0)-8$

$f(0)=-8$

So the y-intercept of the graph is f(0) = -8.

Since the higher degree is 3 , which is odd and the coefficient of higher degree is positive, therefore the ead behavior is defined as,

$f(x)\rightarrow \infty \text{ as }\rightarrow \infty\\f(x)\rightarrow -\infty \text{ as }\rightarrow -\infty$

It means as x decreases unboundedly then the function decreases unboundedly. similarly x increases unboundedly then the function increases unboundedly.

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The intersection can be parameterized by

$C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}$

with $0\le t$ .

By Stoke's theorem, the integral of $\mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k$ along $C$ is equivalent to

$\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS$

where $S$ is the region bounded by $C$ . The line integral reduces to

$\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt$
$=72\pi$