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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

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    $\begingroup$ The acceleration (which is the rate of change in the speed, where the speed is the rate of change in the function's value - in other words acceleration is the rate of change of slope, not the slope itself) is always $2a$ (in particular at the vertex). You can't really point to acceleration on the graph of a position function too easily. The slope of the tangent line at the $y$-intercept is $b$. I am assuming the quadratic function is $ax^2+bx+c$ (keep in mind in math we get to use letters however we want to, so this needs to be specified). $\endgroup$ – anon Sep 4 '13 at 3:18
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    $\begingroup$ The number $b$ controls to what degree the parabola is shifted left or right from the "standard" parabola whose axis of symmetry is the $y$-axis. $\endgroup$ – André Nicolas Sep 4 '13 at 3:19
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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 $a, b, c$.

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  • $\begingroup$ +1 that helps, thanks $\endgroup$ – Emi Matro Sep 6 '13 at 5:25
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We're assuming that the quadratic function is $y(x) = ax^2 + bx + c$, I suppose.

The effect of $c$ is clear -- it just shifts the curve up or down, so let's just set $c=0$ and forget about it, so we just have $y(x) = ax^2 + bx$.

The curve will be a "hump" if $a<0$ and a "hollow" if $a>0$. Let's assume that $a>0$. Then the curve will look like this:

enter image description here

The meaning of $a$ and $b$ can be understood by looking at the points $A$, $B$, $C$, $D$.

The point $A$ is just the origin. The curve passes through the origin because $y(0) = 0$.

The slope at $A$ is $b$ and the slope at $B$ is $-b$. So, as you said, $b$ is a measure of slope.

Now let's think about the triangle $ABD$. Its width is $w = -\frac{b}{a}$, and its height is $h = -\frac{b^2}{2a}$, so we can calculate $\frac{h}{w^2} = -\frac{a}{2}$. So, roughly speaking, $a$ is a measure of the height/width ratio of the triangle.

Another way to look at things, if you know some calculus ... You can show that $\frac{d^2y}{dx^2} = 2a$, so $a$ is also a measure of the second derivative of the curve, and this in turn tells us how fast the slope is changing as we move along the curve. If the slope changes rapidly (large value of $a$), then the triangle $ABD$ will be tall and thin.

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  • $\begingroup$ For some reason uploading my picture isn't working. I'll try again later. $\endgroup$ – bubba Sep 4 '13 at 7:27

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