Help me please !!Algebra Two!!

Mathematics

1 answer:

Mazyrski [523]3 years ago

4 0

Answer:

The zeros of a graph form the factors of the function. When we multiply the factors out and add the exponents up in the final function, we have the highest exponent known as the degree.

Step-by-step explanation:

A polynomial graph has several features we look for to determine the equations.

The zeros of the function are the x-intercepts. If the x-intercepts touch but do not cross then the intercepts have an even multiplicity like 2, 4, 6, etc. If the x-intercepts cross over then they have an odd multiplicity.

Degree is the exponent or multiplicity of each zero. Therefore if we know the multiplicity of each zero we can add them together to find or make an educated guess for the degree of the entire polynomial.

The shape of the graph tells us what type of polynomial. Odd degrees have a backwards S shape. Even degrees have a W shape. The shape can even tell us the if the equation has a positive or negative leading coefficient. Upside down W or an M shape is negative. While a sideways S shape is negative.

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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$