Why does the value of $b$ cause parabola to shift down and left if positive and down and right if negative?

Question

It makes perfect sense why $a$ and $c$ affect the parabola in the ways they do but I don't understand how $b$ has this effect on the graph.

Answer 1

For some intuition, consider this: for the graph $y=ax^2+bx+c$, the coefficient $c$ denotes the height at which the it intersects the $y$-axis. The coefficient $b$, entirely analogously, denotes the slope at which the graph intersects the $y$-axis.

If $b$ is positive and $a$ is positive, then with the intuition from the above paragraph, it makes complete sense that this makes the parabola move down and to the left, compared to if $b=0$.

Answer 2

This is where what you learned about the equation of a parabola and "completing the square" becomes helpful. The "vertex form" for the equation is $$ y \ \ = \ \ ax^2 + bx + c \ \ = \ \ a·\left(x^2 \ + \ \frac{b}{a}·x \ + \ \frac{b^2}{4a^2} \right) \ + \ \left(c \ - \ a·\frac{b^2}{4a^2} \right) $$ $$ = \ \ a·\left(x + \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ \ = \ \ a·(x - h)^2 \ + \ k \ \ , $$ with the vertex of the parabola located at $ \ \left(h \ = \ -\frac{b}{2a} \ , \ k \ = \ c - \frac{b^2}{4a} \right) \ \ . $

For the $ \ x-$coordinate of the vertex then, we can see that when $ \ a > 0 \ \ $ (an "upward-opening" parabola), increasing $ \ b \ $ will cause $ \ h \ $ to decrease , which is a horizontal "shift to the left"; for $ \ a < 0 \ \ $ ("downward-opening"), $ \ b \ $ and $ \ h \ $ have the same sign, so $ \ h \ $ increases as $ \ b \ $ increases (a "shift to the right").

As for the $ \ y-$coordinate, the term $ \ \frac{b^2}{4a} \ $ is always subtracted from $ \ c \ \ . $ For $ \ a > 0 \ \ , $ this term is always positive, so $ \ k \ $ is equal to $ \ c \ $ (the $ \ y-$intercept) when $ \ b = 0 \ $ and always decreases as $ \ b \ $ is either increased or decreased from zero; for $ \ a < 0 \ \ , \ \ \frac{b^2}{4a} \ $ is always negative, so $ \ k \ $ only increases from $ \ c \ $ for such changes.

A question frequently asked in courses around this point is about what the vertex does on a graph when "sliders" are used: for fixed values of $ \ a \ $ and $ \ c \ \ , $ why does the location of the parabola's vertex itself trace out a parabola as $ \ b \ $ is varied? If we take the equation for the $ \ y-$coordinate of the vertex and re-write it as a function of the $ x-$coordinate, we find $$ k \ \ = \ \ c \ - \ \frac{b^2}{4a} \ \ = \ \ c \ - \ a·\frac{b^2}{4a^2} \ \ = \ \ c \ - \ a·\left(\frac{b }{2a} \right)^2 \ \ = \ \ c \ - \ ah^2 \ \ , $$ much of which we saw while completing the square. So this shows that the vertex lies on a "downward-opening" parabola as we vary $ \ h \ $ (or $ \ b \ $ ) when $ \ a > 0 \ $ and traces out an "upward-opening" one for $ \ a < 0 \ \ . $ [This "vertex-tracing" parabola has its vertex on the $ \ y-$axis at $ \ y = c \ \ . \ ] $