SOLUTION: Quadratic Function
Find the a: vertex
b: domain
c: range
of f(x) = -2x^2 - 4x +3
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Question 182550: Quadratic Function
Find the a: vertex
b: domain
c: range
of f(x) = -2x^2 - 4x +3
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Quadratic Function
Find the a: vertex
b: domain
c: range
of f(x) = -2x^2 - 4x +3
----------------
Put in vertex form:
2x^2 + 4x + ? = -y + 3 + ?
2(x^2+2x + 1) = -y + 3 + 2
2(x+1)^2 = -y + 5
(x+1)^2 = (-1/2)y + (5/2)
(x+1)^2 = (-1/2)(y - 5)
----------
Vertex: (-1,5)
domain: all Real Numbers
Range: all Real Numbers <= 5
-----------------------------------
===================================================
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The vertex of the parabola described by
 = ax^2 + bx + c)
is
\right))
So, calculate 
to obtain the x-coordinate of the vertex, then calculate the value of your function at that x-coordinate value to obtain the y-coordinate of the vertex.
If 
then the parabola opens upward and the vertex is a minimum.
If 
then the parabola opens downward and the vertex is a maximum.
Since
where 
is defined for all real x, the domain is:
For a parabola that opens upward, the range is:
\})
For a parabola that opens downward, the range is:
\})
John
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