SOLUTION: Find the vertex, the line of symetry, and the maximum/minimum value of f(x). Graphe the function. f(x)=1/4(x+3)^2+7. Is the value f(-3)=7 a minimum or a maximum

Question 450628: Find the vertex, the line of symetry, and the maximum/minimum value of f(x). Graphe the function. f(x)=1/4(x+3)^2+7. Is the value f(-3)=7 a minimum or a maximum
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Putting the quadratic in standard form gives:
f(x) = 1/4(x^2+6x+9) + 7 = 1/4x^2 + 3/2x + 37/4
The x-value of the vertex is -b/2a = (-3/2)/(2/4) = -3
Substitute this value into the original equation to get the y-value of the vertex:
f(x) = y = 1/4(0)^2 + 7 = 7
So the vertex is (-3,7)
The line of symmetry is the vertical line which passes through the vertex:
x = -3
The parabola opens upward so the maximum value is f(x) =
The minimum value is the y-value of the vertex: f(x) = 7

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