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2 years ago

What is the vertex of the graph of this equation y=2x^2+8x-24

Mathematics

1 answer:

sdas [7]2 years ago

3 0

Answer:

The vertex would be D) (-2, -32)

Step-by-step explanation:

To find the vertex , start by find the x-value. For this we can use the equation below.

x = -b/2a

In which a is the coefficient of x^2 and b is the coefficient of x.

x = -b/2a

x = -(8)/2(2)

x = -8/4

x = -2

Now that we have x, we find y by plugging -2 in for each x.

y = 2x^2 + 8x - 24

y = 2(-2)^2 + 8(-2) - 24

y = 2(4) - 16 - 24

y = 8 - 16 - 24

y = -32

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If (8, −9) and (6, −3) are the endpoints of the diameter of a circle, what is the equation of the circle?

Umnica [9.8K]

$(x-7)^{2} + (y+6)^{2} = 10$ is the equation of the required circle.

Step-by-step explanation:

Step 1 :

Let A = (8,-9) and B = (6,-3) be the endpoints of the diameter of the given circle.

We need to determine the circle's equation with this diameter

Step 2 :

The mid point of the diameter will be center of the required circle.

Let M = (x,y) be the mid point of the circle.

So we have x = (8+6)÷ 2 , y = ((-9) + (-3)) ÷ 2

M = ( 7, -6) is the midpoint of the diameter and the center of the circle.

Step 3 :

The radius of the circle is distance between midpoint and any one of the end point of the diameter that is distance between point A(8,-9) and M ( 7,-6)

Distance between the 2 points $(x_{1} ,y_{1}) and (x_{2} ,y_{2})$ is given by

$\sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2} }$

So distance between A(8,-9) and M ( 7,-6) = $\sqrt{(7-8)^{2}+ (-6-(-9))^{2} }$

= $\sqrt{(1)^{2}+ (3)^{2} } = \sqrt{10 }$

Hence the radius r is $\sqrt{10}$

Step 4 :

The equation of circle with center (a,b) = (7,-6) and radius r = $\sqrt{10}$ is given by

$(x-a)^{2} + (y-b)^{2} = r^{2}$

=> $(x-7)^{2} + (y-(-6))^{2} = (\sqrt{10} )^{2}$

=> $(x-7)^{2} + (y+6)^{2} = 10$ is equation of the required circle.

Step 5 :

Answer :

$(x-7)^{2} + (y+6)^{2} = 10$ is equation of the required circle.

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Step-by-step explanation:

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