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Featured 3 months ago

#"the equation of the parabola is of the form"#

#â€¢color(white)(x)(y-k)^2=4p(x-h)#

#"this parabola opens horizontally"#

#â€¢ " if "4p>0" opens to the right"#

#â€¢ " if "4p<0" opens to the left"#

#"vertex "=(h,k)" and "4p=-12rArrp=-3#

#"here "(h,k)=(-3,2)larrcolor(red)" vertex"#

#"since "4p<0" then opens to the left"#

#"the vertex is midway between the focus and directrix"#

#p=-3" is the distance from the vertex to the focus"#

#rArr"focus "=(-3-3,2)=(-6,2)larrcolor(red)" focus"#

#"the directrix is also 3 units from the vertex in the"#

#"opposite side from the focus"#

#rArrx=0larrcolor(red)" equation of directrix"#

graph{(y-2)^2=-12(x+3) [-10, 10, -5, 5]}

Featured 2 months ago

Set up the long division like this:

#color(magenta)(x^2)+2x+1bar(|"Â Â "color(magenta)(x^4)-2x^3-4x^2+2x+3)#

Divide

#color(white)(x^2+2x+1bar(|"Â Â "color(red)(x^2)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

Multiply

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(blue)(x^4+2x^3+color(white)(1)x^2))#

Subtract

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4"Â Â ")color(orange)(-4x^3-5x^2)"Â Â Â Â Â Â Â Â Â Â ")#

Copy the

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+color(green)(2x)+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2"Â "color(green)(+2x))#

Repeat this process twice, dividing the latest leading term below your line by the leading

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-color(red)(4x))#

#color(magenta)(x^2)+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-color(magenta)(4x^3)-5x^2+2x)#

...

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-color(red)(4x))#

#color(red)(x^2+2x+1)bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar("Â Â Â Â Â "color(blue)(-4x^3-8x^2-4x)))#

...

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-4x)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+color(green)(3))#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar("Â Â Â Â Â "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â ")bar(color(white)(-4x^3+)color(orange)(3x^2+6x)+color(green)(3))#

...

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-4x"Â Â "+color(red)3)#

#color(magenta)(x^2)+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar("Â Â Â Â Â "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â ")bar(color(white)(-4x^3+)color(magenta)(3x^2)+6x+3)#

...

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-4x"Â Â "+color(red)(3))#

#color(red)(x^2+2x+1)bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar("Â Â Â Â Â "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â ")bar(color(white)(-4x^3+)3x^2+6x+3)#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â "bar(color(blue)("Â Â Â Â Â Â Â Â Â Â Â Â "3x^2+6x+3)#

...

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^2-4x"Â Â "+3)#

#x^2+2x+1bar(|"Â Â "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|"Â Â "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar("Â Â "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar("Â Â Â Â Â "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â ")bar(color(white)(-4x^3+)3x^2+6x+3)#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â "bar(color(black)("Â Â Â Â Â Â Â Â Â Â Â Â "3x^2+6x+3)#

#color(white)(x^2+2x+1|"Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ")bar(color(white)(3x^2+6x+color(orange)0)#

Featured 2 months ago

The roots are

Given:

#2x^4-9x^3-7x^2+54x-40=0#

By the rational roots theorem, any *rational* roots of this polynomial equation must be expressible in the form

That means that the only possible *rational* roots are:

#+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-8, +-10, +-20, +-40#

There is a shortcut to trying these in that the sum of the coefficients of the given quartic is zero. That is:

#2-9-7+54-40 = 0#

We can deduce that

#2x^4-9x^3-7x^2+54x-40 = (x-1)(2x^3-7x^2-14x+40)#

Notice that the ratio of the first and second terms of the remaining cubic is different from that of the third and fourth terms. So this cubic will not factor by grouping.

Let's try one of the possible rational roots,

#2(color(blue)(2))^3-7(color(blue)(2))^2-14(color(blue)(2))+40 = 16-28-28+40 = 0#

So

#2x^3-7x^2-14x+40 = (x-2)(2x^2-3x-20)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of

The pair

Use this pair to split the middle term and factor by grouping:

#2x^2-3x-20 = (2x^2-8x)+(5x-20)#

#color(white)(2x^2-3x-20) = 2x(x-4)+5(x-4)#

#color(white)(2x^2-3x-20) = (2x+5)(x-4)#

Hence the remaining zeros are:

#x = -5/2" "# and#" "x = 4#

Featured 2 months ago

Use the fact that a parabola is the locus of points equidistant from the focus point and the directrix line.

The distance from the directrix,

The distance from the focus,

Simplify:

Because the distances must be equal, we can set the right side of equation [1] equal to the right side of equation [2]:

Square both sides:

Expand the squares:

Combine like terms:

Divide both sides by -4:

Featured 1 month ago

See the explanation below

The equation of the hyperbola is

Comparing this equation to your equation

Rearranging your equation

Therefore,

The center of the hyperbola is

The vertices are

The foci are

graph{25x^2-16y^2-1=0 [-1.706, 1.712, -0.853, 0.855]}

Featured 1 month ago

There are three properties of logarithms that will be useful here:

Begin by using the Quotient Property to split apart the logarithm:

Now, rewrite the square root as a fractional exponent:

Now use the Product Property to split apart the first logarithm:

Lastly, use the Exponent Property to rewrite the logarithms:

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