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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...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 $2$ decimal 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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    @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".2012-09-03
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The first steps I would do would be to recall the definition of the image set, and quickly sketch the graph of the function.