SOLUTION: please help me, very lost
Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function.
f(x) = 1/5(x+8)^2+2
1) The vertex is
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: please help me, very lost
Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function.
f(x) = 1/5(x+8)^2+2
1) The vertex is
Log On
Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the function.
f(x) = 1/5(x+8)^2+2
1) The vertex is "ordered pair" ??
2) The line of symmetry is x = ??
3) What is the maximum/minimum value of f(x) ??
4) Is the value, f(-8) = 2 a minimum or maximum ?? Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! Don't be lost, study the "vertex form".
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The equation is already in vertex form, where (h,k) is the vertex.
1) Comparing, (h,k)=(-8,2)
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2) The vertex lies on the axis of symmetry,
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3) Since and , then the parabola opens upwards and the value at the vertex () is a minimum.
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4) From 3), it's a minimum.
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