SOLUTION: Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic fuction, and graph the function.
{{{f(x)=3x^2-12x+16}}}
The vertex is ___ (Type an ordere
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Question 394799: Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic fuction, and graph the function.
The vertex is ___ (Type an ordered pair.)
The line of symmetry is:
A. y=4
B. x=2
C. x=-2
D. y=-4
The maximum/minimum of f(x) is ___.
Is the value, f(2)=4 a minimum or a maximum?
Graph.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic fuction, and graph the function.
f(x)=3x^2-12x+16
---
Vertex occurs at x = -b/(2a) = 12/(6) = 2
f(2) = 3*2^2-12*2+16 = 12-24+16 = 4
The vertex is (2,4) (Type an ordered pair.)
The line of symmetry is x=2
-----------
A. y=4
B. x=2
C. x=-2
D. y=-4
The maximum/minimum of f(x) is (2,4).
Is the value, f(2)=4 a minimum or a maximum?
Minimum
-----------------
Graph.
================
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Given
\ =\ ax^2\ +\ bx\ +\ c)
The 
-coordinate of the vertex is given by:
The 
-coordinate of the vertex is given by:
\ =\ a\left(-\frac{b}{2a}\right)^2\ +\ b\left(-\frac{b}{2a}\right)\ +\ c)
Hence the vertex is:
)
The line of symmetry is:
The maximum or minimum value is )
If 
then the parabola opens upward and 
is a minimum. If 
then the parabola opens upward and is a maximum.
John
My calculator said it, I believe it, that settles it
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