SOLUTION: Please help. I was able to put g(x) = 5x^2 - 30x + 55 into standard form like so g(x) = 5(x - 3)^2+10 but I don't know the value of g or it's max or min Questions here: http://i.

Algebra ->  Functions -> SOLUTION: Please help. I was able to put g(x) = 5x^2 - 30x + 55 into standard form like so g(x) = 5(x - 3)^2+10 but I don't know the value of g or it's max or min Questions here: http://i.      Log On


   



Question 905169: Please help. I was able to put g(x) = 5x^2 - 30x + 55 into standard form like so g(x) = 5(x - 3)^2+10 but I don't know the value of g or it's max or min
Questions here:
http://i.imgur.com/EVIiBCc.png

Found 2 solutions by Math_Boss, Edwin McCravy:
Answer by Math_Boss(45) About Me  (Show Source):
You can put this solution on YOUR website!
g(x) = 5(x − 3)^2+10
Value of g depends on x.
When
x=1 g(1)=5(1 − 3)^2+10=30
x=2 g(2)=5(2 − 3)^2+10=15
x=3 g(3)=5(3 − 3)^2+10=10
At x=3 g(3)=10. g is min.
x=4 g(4)=5(4 − 3)^2+10=15





Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
g(x) = 5(x - 3)^2 + 10
You have to learn about the graph of equations in the form

f(x) = a(x - h)^2 + k, 

[just like you had to learn about the graph of equation in 
the form of y = mx+b, which were straight lines.]

The graph of equations in the form

f(x) = a(x - h)^2 + k

are not lines, but are U-shaped curved graph, a right-side up U
if "a" is a positive number,  and the U is upside-down if "a"
is a negative number.

The vertex is the point (h,k).  The vertex is the bottom of the 
U (a MINIMUM pointr) or the top of the upside-down U, (a MAXIMUM
point) 

Notice that the sign of h is changed but the sign of k is not.

So the graph of your equation

g(x) = 5(x - 3)^2 + 10

has a=5, h=3, and k=10

So since "a" is a positive number 5, the graph is a right side up 
U-shaped graph with the bottom (MINIMUM) point of the U being 
the point (h,k) = (3,10).

Here's the graph



Edwin