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Featured 5 months ago

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(x^3)(x+2)color(magenta)(-2x^3)+5x^3+6x^2-x-2#

#=color(red)(x^3)(x+2)color(red)(+3x^2)(x+2)cancel(color(magenta)(-6x^2))cancel(+6x^2)-x-2#

#=color(red)(x^3)(x+2)color(red)(+3x^2)(x+2)color(red)(-1)(x+2)cancel(color(magenta)(+2))cancel(-2)#

#"quotient "=color(red)(x^3+3x^2-1)," remainder "=0#

#rArr(x^4+5x^3+6x^2-x-2)/(x+2)#

#=(cancel((x+2))(x^3+3x^2-1))/cancel((x+2))=x^3+3x^2-1#

Featured 4 months ago

One should complete the squares so that the equation may be written in one of the two following forms:

where

Given:

Add 40 to both sides:

Group the x terms and the y terms together:

We cannot complete the square unless the leading coefficient is 1, therefore, we remover a factor of 8 from the y terms:

Because

Matching the x terms with the general pattern,

will allow us to solve for the value of h:

This means that

Combine like terms:

We want insert

Matching the y terms with the general pattern,

will allow us to solve for the value of k:

This means that

Combine like terms:

Divide both sides by 64:

Write the denominators as squares:

This is the same form as equation [1],

The center,

The vertices are,

The foci are,

The eccentricity is

Featured 4 months ago

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#

#"is a multiplier"#

#â€¢ " if "a>0" then f(x) is a minimum"#

#â€¢ " if "a<0" then f(x) is a maximum"#

#(h,k)" will be the max/min turning point of f(x)"#

#"to express "f(x)=-1/2x^2-2x+3" in vertex form"#

#"use the method of "color(blue)"completing the square"#

#â€¢ " the coefficient of the "x^2" term must be 1"#

#rArrf(x)=-1/2(x^2+4x)+3#

#â€¢ " add/subtract "(1/2"coefficient of x-term")^2" to"#

#x^2+4x#

#f(x)=-1/2(x^2+2(2)xcolor(red)(+4)color(red)(-4))+3#

#color(white)(f(x))=-1/2(x+2)^2+2+3#

#color(white)(f(x))=-1/2(x+2)^2+5larrcolor(blue)"in vertex form"#

#rArrcolor(magenta)"vertex "=(-2,5)" and "a<0#

#rArr"maximum turning point "=(-2,5)#

graph{(y+1/2x^2+2x-3)((x+2)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

Featured 3 months ago

The function is

As the denominator must be

The domain is

To find the range, proceed as follows :

In order for this quadratic equation in

Therefore,

The range is

graph{x/(x^2-4) [-10, 10, -5, 5]}

Featured 2 months ago

Recall that, for the **Hyperbola**

asymptotes are given by,

In our **Case,** the asymptotes are,

Next,

Multiplying by

By

With **Hyperbola** is,

Featured 2 months ago

Perform the Gauss Jordan elimination on the augmented matrix

I have written the equations not in the sequence as in the question in order to get

Perform the folowing operations on the rows of the matrix

Thus

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