SOLUTION: Find the maximum or minimum value of the function: f(x) = 1 + 3x -x^2 = x^2 + 3x + 1 =-(x^2 -3x + 2.25) +1 + 2.25 =-(x^2 -3x + 2.25) + 3.25 x= -b/2a = -3/2(-1) = 3/2 = max

Algebra ->  Functions -> SOLUTION: Find the maximum or minimum value of the function: f(x) = 1 + 3x -x^2 = x^2 + 3x + 1 =-(x^2 -3x + 2.25) +1 + 2.25 =-(x^2 -3x + 2.25) + 3.25 x= -b/2a = -3/2(-1) = 3/2 = max      Log On


   

Question 45202This question is from textbook College Algebra
: Find the maximum or minimum value of the function:
f(x) = 1 + 3x -x^2
= x^2 + 3x + 1
=-(x^2 -3x + 2.25) +1 + 2.25
=-(x^2 -3x + 2.25) + 3.25
x= -b/2a = -3/2(-1) = 3/2 = max 1.5
y = 3.25
Thank you so much! This question is from textbook College Algebra

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This is also correct. Again, if you use the formula x=+-b%2F%282a%29 to find the x coordinate of the vertex, then it's easier to substitute this value of x into the original equation to find f(x).

R^2 at SCC