SOLUTION: How do I find the Ymax, Ymin, Xmax, Xmin? At the moment I am trying to find the Ymax of y=-x^2-4x-3 Thank you

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do I find the Ymax, Ymin, Xmax, Xmin? At the moment I am trying to find the Ymax of y=-x^2-4x-3 Thank you      Log On


   

Question 951490: How do I find the Ymax, Ymin, Xmax, Xmin?
At the moment I am trying to find the Ymax of y=-x^2-4x-3
Thank you
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39628) About Me  (Show Source):
You can put this solution on YOUR website!
How do you mean for those requested values?

y=-x%5E2-4x-3
-%28x%5E2%2B4x%2B3%29
-%28x%5E2%2B4x%2B4%2B3-4%29
-%28%28x%2B2%29%5E2-1%29
highlight_green%28y=-%28x%2B2%29%5E2%2B1%29

The domain is all real numbers.
The range is y%3C=1.

Xmax and Xmin are meaningless.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
y=-x%5E2-4x-3+
here, you have a parabola and max or min is at vertex
write it in a vertex form y=a%28x-h%29%2Bk where h and k are x and ycoordinates of the vertex
y=a%28x-h%29%2Bk ...if a%3C0 parabola opens downward and has a max, and if a%3E0 parabola opens upward and has a min
y=-x%5E2-4x-3+....complete square
y=-%28x%5E2%2B4x%29-3+
y=-%28x%5E2%2B4x%2B_%29-_-3+
y=-1%2A%28x%5E2%2B4x%2B2%5E2%29-%28-1%292%5E2-3+
y=-%28x%2B2%29%5E2%2B4-3+
y=-%28x%2B2%29%5E2%2B1+
so, h=-2 and k=1 and vertex is at (-2,1)
then, max is at x=-2 where y=1