SOLUTION: I am supposed to find the maximum value of this equation and am not having any
luck f(x)= -10/3(x+5)^2 - 5 This is what I have y= -10(x+5)^2/3 - 5*3/3, y=
(-10(x+5)^2)/3 - 15/3
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-> SOLUTION: I am supposed to find the maximum value of this equation and am not having any
luck f(x)= -10/3(x+5)^2 - 5 This is what I have y= -10(x+5)^2/3 - 5*3/3, y=
(-10(x+5)^2)/3 - 15/3
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Question 337182: I am supposed to find the maximum value of this equation and am not having any
luck f(x)= -10/3(x+5)^2 - 5 This is what I have y= -10(x+5)^2/3 - 5*3/3, y=
(-10(x+5)^2)/3 - 15/3, -10x^2-100x-265/3, factor out -5, y= (-5(2x^2 +
20x+53)/3, this is where I end. Please help me understand Thank-you.
I think the answer is -5 as the constant - 5 = k in the equation f(x)=a(x-h)^2+k Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! No, the function is already in vertex form.
The min or max value occurs at the vertex.
Since the coefficient of the term is negative, the parabola opens downwards and the max value is
.
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