SOLUTION: Consider the quadratic function f(x)=3x^2+66x+311 (a)Find the vertex of f and then write the formula Find vertex form as f(x)=a(x-h)^2+k (b)what the range of f?

Question 424981: Consider the quadratic function f(x)=3x^2+66x+311
(a)Find the vertex of f and then write the formula Find vertex form as f(x)=a(x-h)^2+k
(b)what the range of f?
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
(a) The vertex is the turning point of the function in which df(x)/dx = 0
Taking the derivative and setting=0 we get
df/dx = 0 = 6x + 66 -> x = -11
Sustitute this value into the original equation:
f(-11) = 3(-11)^2 + 66(-11) + 311 = 363 - 726 + 311 = -52
So the vertex (h,k) = (-11,-52)
Writing in vertex form we have f(x) = 3(x+11)^2 - 52

(b) The range of the function is the set of all possible values of f(x)
So the range is [k,]

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