Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.
f(x) = −one twelfth(x − 5)2 + 2
f(x) = one twelfth(x − 5)2 + 2
f(x) = −one twelfth(x + 5)2 + 2
f(x) = one twelfth(x + 5)2 + 2

Answers

Answer 1

Answer:

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is [tex]y = \frac{1}{12}\cdot (x+5)^{2}+2[/tex].

Step-by-step explanation:

From statement we understand that parabola has its axis of symmetry in an axis parallel to y-axis. According to Analytical Geometry, the minimum distance between focus and directrix equals to twice the distance between vertex and any of endpoints.

If endpoints are (-5, 5) and (-5, -1), respectively, then such distance ([tex]r[/tex]), dimensionless, is calculated by means of the Pythagorean Theorem:

[tex]r = \frac{1}{2}\cdot \sqrt{[-5-(-5)]^{2}+[5-(-1)]^{2}}[/tex]

[tex]r = 3[/tex]

And the location of the vertex ([tex]V(x,y)[/tex]), dimensionless, which is below the focus, is:

[tex]V(x,y) = F(x,y)-R(x,y)[/tex] (1)

Where:

[tex]F(x,y)[/tex] - Focus, dimensionless.

[tex]R(x,y)[/tex] - Vector distance, dimensionless.

If we know that [tex]F(x,y) = (-5,5)[/tex] and [tex]R(x,y) = (0,3)[/tex], then the location of the vertex is:

[tex]V(x,y) = (-5,5)-(0,3)[/tex]

[tex]V(x,y) =(-5,2)[/tex]

In addition, we define a parabola by the following expression:

[tex]y-k = \frac{(x-h)^{2}}{4\cdot r}[/tex] (2)

Where:

[tex]h[/tex], [tex]k[/tex] - Coordinates of the vertex, dimensionless.

[tex]r[/tex] - Distance of the focus with respect to vertex, dimensionless.

If we know that [tex]h = -5[/tex], [tex]k = 2[/tex] and [tex]r = 3[/tex], then the equation of the parabola is:

[tex]y = \frac{1}{12}\cdot (x+5)^{2}+2[/tex]

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is [tex]y = \frac{1}{12}\cdot (x+5)^{2}+2[/tex].


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Step-by-step explanation:

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Step-by-step explanation:

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In testing the validity of a multiple regression model in which there are four independent variables, the null hypothesis is H0 : β1 = β2 = β3 = β4 = 0.

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

In a certain country license plates consist of zero or one digit followed by four or five uppercase letters from the Roman alphabet.
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Answer:

Step-by-step explanation:

From the given information:

The number of different license plate the country  can produce is in the form:

=L₁L₂L₃L₄   or  L₁L₂L₃L₄L₅  or  D₁L₁L₂L₃L₄   or    D₁L₁L₂L₃L₄L₅

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A) The country can produce 118,813,760 different licence plates.

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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1. (01.02)
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Answers

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Step-by-step explanation:

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Identify the y-intercept and zeros of
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Answers

Answer:

Our y-intercept is (0, -36).

Our real zeros are: (-2√3, 0) and (2√3, 0)

And our imaginary zeros are: (i√3, 0) and (-i√3, 0)

Step-by-step explanation:

We have the function:

[tex]y=x^4-9x^2-36[/tex]

Let's identify the y- and x-intercepts.

Y-Intercept:

To determine the y-intercept, we will substitute 0 for x and solve for y. So:

[tex]y=(0)^4-9(0)^2-36[/tex]

Evaluate:

[tex]y=-36[/tex]

Therefore, our y-intercept is at (0, -36).

X-Intercepts:

To determine our x-intercepts, we need to substitute 0 for y and solve for x. So:

[tex]0=x^4-9x^2-36[/tex]

Now, we can factor. Before doing so, we can make the factored easier to see by converting this into quadratic form. Let's let u=x². Then:

[tex]0=(u^2)^2-9(x^2)-36[/tex]

Substitute:

[tex]0=u^2-9u-36[/tex]

Now, let's factor. We can use -12 and 3. So:

[tex]0=u^2+3u-12u-36 \\ 0=(u(u+3)-12(u+3)) \\ 0=(u-12)(u+3)[/tex]

Now, we can substitute back u:

[tex]0=(x^2-12)(x^2+3)[/tex]

Zero Product Property:

[tex]x^2-12=0\text{ or } x^2+3=0[/tex]

Solve for x in each case:

[tex]x^2=12 \text{ or } x^2=-3[/tex]

Take the square root of both sides. Since we're taking an even root, we will need plus/minus. Therefore:

[tex]x=\pm\sqrt{12}=\pm2\sqrt{3}\text{ or } x=\pm\sqrt{-3}=\pm i\sqrt{3}[/tex]

Therefore, our zeros (both real and imaginary) are: (2√3, 0), (-2√3, 0), (i√3, 0), and (-i√3, 0)

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