SOLUTION: please help me answer....y=(x-4)^2+6...also need to know minimum and maximum points, the equation of the line of symmetry, the distance and direction of the transition of the parab
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-> SOLUTION: please help me answer....y=(x-4)^2+6...also need to know minimum and maximum points, the equation of the line of symmetry, the distance and direction of the transition of the parab
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Question 440799: please help me answer....y=(x-4)^2+6...also need to know minimum and maximum points, the equation of the line of symmetry, the distance and direction of the transition of the parabola as well as whether the parabola opens upward and downward Answer by ewatrrr(24785) (Show Source):
Hi
Using the vertex form of a parabola, where(h,k) is the vertex
y = (x-4)^2+6
1) parabola opens upward a = 1 > 0 Vertex Pt(4,6) is a minumum pt
2) Line of symetry is the x = 4
3) relative to the the translation form the origin: to the right 4 & up 6
