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Given a quadratic, $ax^2+bx+c$. I know c is the intercept, and the sign of $a$ tells us wether it is a positive "u" shape, or negative, an upside down u. But what about b? Is my following observation correct;

If b>0 the min or max is to the left of the y axis If b<0 the min or max is to the right of the y axis? Are these statements true?

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    Every one should know the extremum is attained at $x=-b/2a$. – 2017-01-24
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    Try to think of the interplay between $a$ and $b$ regarding the position of min/max. But your observation shows you are in a good path. – 2017-01-24
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    @Bernard Well, not *everyone*..He might be a school student.. – 2017-01-24
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    @MathematianByMistake Nope, just a rusty old timer getting back in the game :) – 2017-01-24
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    Precisely, this should be learnt in high school. – 2017-01-24
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    @MathsWiz Ha! Even better then! – 2017-01-24
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    That's the way it is here. – 2017-01-24
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    If you think of $c$ as the y-intercept, you can also think of $b$ as the slope of the curve at that y-intercept. This helps a bit with imagining the graph. So $b$ definitely affects the position of the min/max, but it alone can't determine it. – 2017-01-24
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    Welp, now I have a Frank Sinatra song stuck in my head. "Do - b - do - b - do...." Thanks a ton, OP. – 2017-01-24

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You can play with $a, b, c$ here:

http://www.mathwarehouse.com/quadratic/parabola/interactive-parabola.php

Generally, the the coordinates of vertex are $T(-\frac{b}{2a}, \frac{4ac-b^2}{4a})$.

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    I have been playing on google, but are my observations not true? – 2017-01-24
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    @MathsWiz Almost, since the $x$ coordinate of the vertex depends on both $a$ and $b$ you can't just say that it will be on the right/left based on if b>0 or b<0. I can say that your observation is true, but it's the opposite if a is negative. – 2017-01-24
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It is true if $a \gt 0$. You can write it as $a(x+\frac {b}{2a})^2+c-\frac {b^2}{4a}$ The vertex is then at $x=-\frac b{2a}$

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If $b=0$ the graph of $y=ax^2+bx+c$ is symmetric with respect to the $y$ axis, if $b \ne 0$ it is symmetric with respect to the stright line $x=\frac{-b}{2a}$.

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    Why distinguishing the two cases? I see no reason to. – 2017-01-24