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Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function. See the attachment.
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its the maximum, differentiate the function twice, the result turns out to be negative, the condition for maxima(along with dy/dx=0)
ok so...
When the leading coefficient (coef. of x^2) of a parabola is negative, it is a maximum, and vice versa.
ok so did i pick the right one? and did i pick the right graph?
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yes, you did
GREAT THX!!!
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