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I am revising some old exam question that i have now reflecting back on it.. the question is to find the image set of the function.. I forgot how to do this.. can some one tell me the steps not just the answer.. tks.. cause if I remember correctly we need to have a limit right? $$ f(x) = 3x^2 + 12x + 180 $$ on the side note it does mention to give it to 2decimal values

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In the $Oxy$ plane, a curve like $y=ax^2+bx+c$ is a parabola. It is well known that the range of such a function is the half-line $[y_V,+\infty)$ is $a>0$ and $(-\infty,y_V]$ if $a<0$, where $y_V$ is the value of the function at its vertex $x_V=-b/(2a)$. More explicitly, $$ y_V=\frac{4ac-b^2}{4a}. $$

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so u r suggesting there is indeed a limit right? – JackyBoi Sep 3 '12 at 14:23
I'm sorry: what limit? – Siminore Sep 3 '12 at 14:25
i mean to say, quote ur statment 'It is well known that the range of such a function is the half-line [yV,+∞) ' so there is a range right? – JackyBoi Sep 3 '12 at 14:28
Well, any function has a range. If $f \colon A \to B$, then the range of $f$ is by definition $f(A) = \{f(a)\in B \mid a \in A\}$. – Siminore Sep 3 '12 at 14:29
@JackyBoi: Your keyboard is out of order; it seems to be omitting random letters in your comments, and the shift key has entirely stopped working. Please proofread before pressing "Add Comment". – Henning Makholm Sep 3 '12 at 14:46

The first steps I would do would be to recall the definition of the image set, and quickly sketch the graph of the function.

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