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

Question

1.does the function have a minimum or maximum value?

2. what is the functions minimum or maximum value?

3. where does the minimum or maximum occur?

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

Answer 1

g(x) = 2x² - 12x +22 is a parabola. Since the coefficient on x² is positive it opens up and thus has a minimum atthe vertex. To find the vertex completion of the square is used.

 

g(x) = 2(x² - 6x        ) + 22

g(x) = 2(x² - 6x + 9) + 22 - 18

g(x) = 2(x - 3)² + 4

 

The minimum is at x = 3 and the value is 4.

Answer 2

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

The coefficient on the x^2 term is positive (2), meaning that the function has a minimum value. I would find the vertex first, using x = -b/2a = -(-12)/2(2) = 12/4 = 3. Then you can plug that value back into the original expression to get

g(3) = 2(3)^2 – 12(3) + 22 = 4.

What that means is that the minimum value is 4, and it occurs at the x-coordinate of x = 3

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 3

Hi Missy,

 

I'm guessing that the function you're given is actually:

 

    g(x) = 2x² - 12x + 22

 

1.  We know that any "quadratic equation" (one that has a "squared" as the highest exponent) always forms a parabola when we graph it. If the leading "a" coefficient" is positive, that means the parabola opens upward, making a "u" shape, which means that the function must have a minimum - the bottom of the "u". If the leading "a" coefficient is negative, that means the parabola opens downward, making an upside-down "u", and the function instead has a maximum, at the top of the upside-down "u". In this case, the "a" coefficient is 2, so the function must be an upward-opening parabola, and the function must have a minimum.

 

At this point, question #3 is actually easier to answer, and it will help us answer #2. The minimum/maximum of a parabola always occurs at the "vertex", and the x-coordinate of the vertex can always be found by:

 

    x_vertex = -b/(2a)      where "a" is the number in front of the "x squared" term, and "b" is the number in front of the "x" term:

 

    x_vertex = -(-12)/(2*2)  =  12/4  =  3

 

So the minimum occurs when x=3 (answer for #3)

 

For part #2, we want to know the value of the function at the minimum. We now know that the minimum occurs when x=3, so we can simply plug in 3 for the x value in the function:

 

    g(3) = 2*(3)² - 12*(3) + 22

 

    g(3) = 18 - 36 + 22

 

    g(3) = 4

 

The function reaches a minimum value of 4, when x=3.

 

I hope this helps!

Answer 4

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

 

I am assuming you forgot the x in the middle term.   

 

For an equation 

f(x)=ax**2+bx+c

 

the extrema ( either a min or a max) occurs at

 

x=-b/(2a)

 

so for your problem

 

x=12/4=3

 

now if a>0 this is a minimum

       if a<0 this is a maximum

 

in your example a is positive so g(3)=4 is a minimum