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## How do you sketch the parabola (y-2)^2=-12(x+3) and find the vertex, focus, and directrix?

Jim G.
Featured 3 months ago

$\text{see explanation}$

#### Explanation:

$\text{the equation of the parabola is of the form}$

•color(white)(x)(y-k)^2=4p(x-h)

$\text{this parabola opens horizontally}$

• " if "4p>0" opens to the right"

• " if "4p<0" opens to the left"

$\text{vertex "=(h,k)" and } 4 p = - 12 \Rightarrow p = - 3$

$\text{here "(h,k)=(-3,2)larrcolor(red)" vertex}$

$\text{since "4p<0" then opens to the left}$

$\text{the vertex is midway between the focus and directrix}$

$p = - 3 \text{ is the distance from the vertex to the focus}$

$\Rightarrow \text{focus "=(-3-3,2)=(-6,2)larrcolor(red)" focus}$

$\text{the directrix is also 3 units from the vertex in the}$
$\text{opposite side from the focus}$

$\Rightarrow x = 0 \leftarrow \textcolor{red}{\text{ equation of directrix}}$
graph{(y-2)^2=-12(x+3) [-10, 10, -5, 5]}

## X^4-2x-4x^2+2x+3 is divided by x^2+2x+1 Using in long division, find the quotients And the remainder when: can u please calculate for me ?.thank u

Geoff K.
Featured 2 months ago

$\left({x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3\right) \div \left({x}^{2} + 2 x + 1\right) = {x}^{2} - 4 x + 3$, with no remainder.

#### Explanation:

Set up the long division like this:

$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } \textcolor{m a \ge n t a}{{x}^{4}} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$

Divide $\textcolor{m a \ge n t a}{{x}^{4} / {x}^{2}}$, giving $\textcolor{red}{{x}^{2}}$; put this quotient above the ${x}^{4}$:

color(white)(x^2+2x+1bar(|"  "color(red)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$

Multiply $\textcolor{red}{{x}^{2}} \times \left({x}^{2} + 2 x + 1\right)$, giving $\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}$; put this product below the ${x}^{4} - 2 {x}^{3} - 4 {x}^{2}$:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(blue)(x^4+2x^3+color(white)(1)x^2))

Subtract $\left({x}^{4} - 2 {x}^{3} - 4 {x}^{2}\right) - \left(\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}\right)$, giving color(orange)(–4x^3-5x^2); draw a line under $\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}$ and write this difference below the line:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ "color(white)(x^4" ")color(orange)(-4x^3-5x^2)" }}$

Copy the $\textcolor{g r e e n}{2 x}$ from the dividend down below this line:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + \textcolor{g r e e n}{2 x} + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ "color(white)(x^4)-4x^3-5x^2" } \textcolor{g r e e n}{+ 2 x}}$

Repeat this process twice, dividing the latest leading term below your line by the leading ${x}^{2}$ from the divisor:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))
$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - \textcolor{m a \ge n t a}{4 {x}^{3}} - 5 {x}^{2} + 2 x}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))
$\textcolor{red}{{x}^{2} + 2 x + 1} \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l u e}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + \textcolor{g r e e n}{3}}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} \textcolor{\mathmr{and} a n \ge}{3 {x}^{2} + 6 x} + \textcolor{g r e e n}{3}}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)3)
$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} \textcolor{m a \ge n t a}{3 {x}^{2}} + 6 x + 3}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)(3))
$\textcolor{red}{{x}^{2} + 2 x + 1} \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} 3 {x}^{2} + 6 x + 3}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ "bar(color(blue)(" } 3 {x}^{2} + 6 x + 3}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+3)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} 3 {x}^{2} + 6 x + 3}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ "bar(color(black)(" } 3 {x}^{2} + 6 x + 3}$
color(white)(x^2+2x+1|"                       ")bar(color(white)(3x^2+6x+color(orange)0)

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

George C.
Featured 2 months ago

The roots are $1 , 2 , 4 , - \frac{5}{2}$

#### Explanation:

Given:

$2 {x}^{4} - 9 {x}^{3} - 7 {x}^{2} + 54 x - 40 = 0$

By the rational roots theorem, any rational roots of this polynomial equation must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 40$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational roots are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm \frac{5}{2} , \pm 4 , \pm 5 , \pm 8 , \pm 10 , \pm 20 , \pm 40$

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 $x = 1$ is a root and $\left(x - 1\right)$ a factor:

$2 {x}^{4} - 9 {x}^{3} - 7 {x}^{2} + 54 x - 40 = \left(x - 1\right) \left(2 {x}^{3} - 7 {x}^{2} - 14 x + 40\right)$

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, $2$:

$2 {\left(\textcolor{b l u e}{2}\right)}^{3} - 7 {\left(\textcolor{b l u e}{2}\right)}^{2} - 14 \left(\textcolor{b l u e}{2}\right) + 40 = 16 - 28 - 28 + 40 = 0$

So $x = 2$ is a root and $\left(x - 2\right)$ a factor:

$2 {x}^{3} - 7 {x}^{2} - 14 x + 40 = \left(x - 2\right) \left(2 {x}^{2} - 3 x - 20\right)$

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $A C = 2 \cdot 20 = 40$ which differ by $B = 3$.

The pair $8 , 5$ works.

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

$2 {x}^{2} - 3 x - 20 = \left(2 {x}^{2} - 8 x\right) + \left(5 x - 20\right)$

$\textcolor{w h i t e}{2 {x}^{2} - 3 x - 20} = 2 x \left(x - 4\right) + 5 \left(x - 4\right)$

$\textcolor{w h i t e}{2 {x}^{2} - 3 x - 20} = \left(2 x + 5\right) \left(x - 4\right)$

Hence the remaining zeros are:

$x = - \frac{5}{2} \text{ }$ and $\text{ } x = 4$

## How do you find the standard form of the equation Vertex: ( -1, 2), Focus (-1, 0) Vertex: (-2, 1), Directrix: x = 1?

Douglas K.
Featured 2 months ago

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

#### Explanation:

The distance from the directrix, $x = 1$, to any point, $\left(x , y\right)$, on the parabola is:

$d = x - 1 \text{ [1]}$

The distance from the focus, $\left(- 1 , 0\right)$ to any point, $\left(x , y\right)$, on the parabola is:

$d = \sqrt{{\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 0\right)}^{2}}$

Simplify:

$d = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}} \text{ [2]}$

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

$x - 1 = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$

Square both sides:

${\left(x - 1\right)}^{2} = {\left(x + 1\right)}^{2} + {y}^{2}$

Expand the squares:

${x}^{2} - 2 x + 1 = {x}^{2} + 2 x + 1 + {y}^{2}$

Combine like terms:

$- 4 x = {y}^{2}$

Divide both sides by -4:

$x = - \frac{1}{4} {y}^{2} \leftarrow$ standard form for a parabola that opens left.

Featured 1 month ago

See the explanation below

#### Explanation:

The equation of the hyperbola is

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$

Comparing this equation to your equation

$25 {x}^{2} - 16 {y}^{2} - 1 = 0$

$25 {x}^{2} - 16 {y}^{2} = 1$

${x}^{2} / {\left(\frac{1}{5}\right)}^{2} - {y}^{2} / {\left(\frac{1}{4}\right)}^{2} = 1$

Therefore,

$a = \frac{1}{5}$

$b = \frac{1}{4}$

$c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{\frac{1}{25} + \frac{1}{16}} = \pm \frac{\sqrt{41}}{20}$

The center of the hyperbola is $C = \left(0 , 0\right)$

The vertices are $A = \left(\frac{1}{5} , 0\right)$ and $A ' = \left(- \frac{1}{5} , 0\right)$

The foci are $F = \left(\frac{\sqrt{41}}{20} , 0\right)$ and $F ' = \left(- \frac{\sqrt{41}}{20} , 0\right)$

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

## How do you use the properties of logarithms to expand ln((x^4sqrty)/z^5)?

Jason K.
Featured 1 month ago

$4 \ln x + \frac{1}{2} \ln y - 5 \ln z$

#### Explanation:

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

log_a (b/c) = log_a b - log_a c color(white)("aaaaaa")"Quotient Property"

log_a (b*c) = log_a b + log_a c color(white)("aaaaaa")"Product Property"

log_a (b^c) = c*log_a b color(white)("aaaaaaaaaaaaa")"Exponent Property"

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

$\ln \left(\frac{{x}^{4} \sqrt{y}}{{z}^{5}}\right) = \ln \left({x}^{4} \sqrt{y}\right) - \ln \left({z}^{5}\right)$

Now, rewrite the square root as a fractional exponent:

$\ln \left({x}^{4} \sqrt{y}\right) - \ln \left({z}^{5}\right) = \ln \left({x}^{4} \cdot {y}^{\frac{1}{2}}\right) - \ln \left({z}^{5}\right)$

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

$\ln \left({x}^{4} \cdot {y}^{\frac{1}{2}}\right) - \ln \left({z}^{5}\right) = \ln {x}^{4} + \ln {y}^{\frac{1}{2}} - \ln {z}^{5}$

Lastly, use the Exponent Property to rewrite the logarithms:

$\ln {x}^{4} + \ln {y}^{\frac{1}{2}} - \ln {z}^{5} = 4 \ln x + \frac{1}{2} \ln y - 5 \ln z$

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