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3 years ago

2m^2+2m-12=0 in vertex form

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

5 0

Answer:

y = 2(m + 1/2)^ -25/2

Step-by-step explanation:

vertex form is:

y = 2m^2 + 2m - 12

y + 12 + ( )___ =( ) 2m^2 + 2m + ___

Because a is not one, you must divide the entire right side by the a value, which is 2. Then place the 2 in the parenthesis.

y + 12 + (2)__ = (2) m^2 + m + __

Take half of b, b being one.

Half of be is 1/2. Now square it to get 1/4. Place that value in the blanks.

y + 12 + (2)1/4 = (2) m^2 + m + 1/4

Now simplify the left side and factor the right side.

y+ 12 + 1/2 = 2(m + 1/2)^2

y + 25/2 = 2(m + 1/2)^2

Subtract 25/2 from both sides.

y = 2(m + 1/2)^ -25/2

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Read 2 more answers

Identify the standard form of the equation by completing the square.

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Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

<u>Given equation</u>:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$