g(x)=2x^2-12+22

Question

Answer the questions below about the quadratic function.

Answer 1

Hello, thank you for taking the time to post your question!

Did part of the question get chopped off? Usually on this type of question it would be asking for you to solve for things like the zeroes or the extreme values. If it’s wanting you to find the min/max you could first find the vertex using x = -b/2a then plug that value back into g(x) to find the extreme value.

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

I am guessing you are trying to solve for zero.

If you meant the equation
g(x)=2x²-12x+22

Than use the quadratic equation

x= (-b+ SquareRoot(b²-4ac))/2a

The 'a' in the equation is the constant before x², the 'b' is the constant before x, and the 'c' is the final constant.

so a= 2 b=-12 and c =22

plug it in:

x=(-(-12)+ SquareRoot((-12)²-4(2)(22)))/2(2)

simplify

x=(12+ SquareRoot (144-176))/4

simplify further

x=(12+ SquareRoot(-32)) /4

Answers:

x= 3+i SquareRoot(32)/4

or

x=3- i SquareRoot(32)/4

both of which are imaginary numbers

 

 

 

 

If the equation you meant to input was: g(x)=2x²-12+22 

First, combine the constants. 0 = 2x²+(22-12) = 2x²+20

Now, try to get x alone by bringing the constants over. -20 = 2x²

Then, divide both sides by 2. -20/2 = 2x²/2  =>   -10 = x²  

Finally, square root each side to get x alone. (-10)^1/2= (x²)^1/2  

Final answer: x = i(10)^1/2 where i is the imaginary number and equals the square root of negative one.

 

If you meant the equation

g(x)=2x²-12x+22