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2 years ago

NO LINKS!!! Find the equation of the PARABOLA with a vertex at (-2, 6) and passing through the point (1, -3)

Mathematics

2 answers:

levacccp [35]2 years ago

7 0

General equation of parabola

y=a(x-h)²+k

for vertex(h,k)

Now

$\\ \rm\Rrightarrow y=a(x+2)^2+6$

Put (1,-3) and find a

$\\ \rm\Rrightarrow -3=a(1+2)^2+6$

$\\ \rm\Rrightarrow -9=a(3)^2$

$\\ \rm\Rrightarrow -9=9a$

$\\ \rm\Rrightarrow a=-1$

equation of parabola

$\\ \rm\Rrightarrow y=-(x+2)^2+6$

Mrac [35]2 years ago

4 0

Answer:

$y=-(x+2)^2+6$

Step-by-step explanation:

Vertex form of a quadratic equation

$y=a(x-h)^2+k$

where:

(h, k) is the vertex
a is some constant

Given:

vertex = (-2, 6)
point on parabola = (1, -3)

Substitute the given values into the vertex equation and solve for a:

$\implies -3=a(1-(-2)^2+6$

$\implies -3=a(3)^2+6$

$\implies -3=9a+6$

$\implies 9a=-9$

$\implies a=-1$

Vertex form

Substitute the given vertex and the found value of a into the vertex equation:

$\implies y=-(x+2)^2+6$

Standard form

Expand the brackets of the vertex form:

$\implies y=-(x^2+4x+4)+6$

$\implies y=-x^2-4x+2$

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3 years ago

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Moreover, total essays wrote this year are 200.

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3 years ago

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Firlakuza [10]

Answer:

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3 years ago

The lengths of the sides of a triangle are consecutive integers and the largest angle is twice the smallest angle. Find the meas

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Answer:

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Step-by-step explanation:

I could not think of an easy way to solve this, apart from having a graphing calculator do it. In the end, I found I could solve it analytically using a combination of the law of sines and the law of cosines.

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sin(θ)/x = sin(2θ)/(x+2)

Cross-multiplying and using the trig identity for sin(2θ), we have ...

(x +2)sin(θ) = 2x·sin(θ)cos(θ)

Dividing out sin(θ), we see that ...

cos(θ) = (x+2)/(2x)

___

The law of cosines for the shortest side and smallest angle tells you ...

x^2 = (x+1)^2 + (x+2)^2 - 2(x+1)(x+2)·cos(θ)

Substituting the above expression for cos(θ), this can be rewritten as ...

0 = (x^2 +2x +1) +(x^2 +4x +4) -x^2 -(x+1)(x+2)^2/x

0 = x^2 +6x +5 -(x+1)(x+2)^2/x . . . . . . collect terms outside the fraction

0 = x(x+5)(x+1) -(x+1)(x+2)^2 . . . . . . . . factor and multiply by x

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0 = x -4 . . . . . . . . . . . . . . . . . . . . collect terms

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θ = arccos(3/4) ≈ 41.40962° ≈ 41°

_____

The attachment shows a triangle-solver's result using the consecutive integers for side lengths. It confirms the answer we have here.

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3 years ago

NO LINKS!!! Find the equation of the PARABOLA with a vertex at (-2, 6) and passing through the point (1, -3)​

NO LINKS!!! Find the equation of the PARABOLA with a vertex at (-2, 6) and passing through the point (1, -3)