In a quadratic function:
coefficient $a$ controls the speed of increase/decrease from the vertex.
coefficient $b$ controls the downward slope as the function crosses the y-axis.
I don't really get what the coefficient $b$ is doing. Is it just controlling the steepness of the slope after it crosses the y-axis? Isn't that what $a$ is doing by controlling the speed of increase/decrease (i.e., controlling the steepness/slope of the graph and how "closed" it is)?
Why should the slope change as it crosses the y-axis?
I guess I'm not really seeing the difference between $a$ and $b$.
If possible, in addition to an explanation, could you provide a picture of a parabola (quadratic function) and point to what part of the graph is being determined by $a$ and $b$?
Thank you
Best Answer
It is convenient to rewrite $$ f(x) = a x^{2} + b c + c = a \left(x + \frac{b}{2 a}\right)^{2} + \frac{4 a c - b^{2}}{4 a}, $$ as in the solution of the corresponding quadratic equation. (I assume $a \ne 0$.)
Now you may think to start with the graph of $$f_{1}(x) = x^{2},$$ which I suppose you understand.
Then consider $$f_{2}(x) = \left(x + \dfrac{b}{2 a}\right)^{2}.$$ You have shifted the graph left or right by the absolute value of $\frac{b}{2 a}$.
Then consider $$f_{3}(x) = \left(x + \dfrac{b}{2 a}\right)^{2} + \dfrac{4 a c - b^{2}}{4 a}.$$ You have shifted the graph up or down by the absolute value of $\dfrac{4 a c - b^{2}}{4 a}$.
Finally you get $$ f(x) = a \left(x + \frac{b}{2 a}\right)^{2} + \frac{4 a c - b^{2}}{4 a}, $$ and the effect of $a$ is to stretch or shrink the graph horizontally (depending on the absolute value of $a$ being greater than, or less than, $1$), possibly turning it upside down, if $a$ is negative.
So the reason you were puzzled is that what really counts are some algebraic combinations of the parameters $a, b, c$.