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

Algebra ->  Rational-functions -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
No, the function is already in vertex form.
a=-10%2F3
h=-5
k=-5
The min or max value occurs at the vertex.
Since the coefficient of the x%5E2 term is negative, the parabola opens downwards and the max value is
ymax=k=-5
.
.
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