The standard equation is ax^2 + bx + c. While I was plotting the graph on Desmos, an increase in the value of 'a' would cause the graph to be narrow, and a decrease would cause it to be broad. As for 'b', it would cause the graph to shift sideways/horizontally; for 'c', it would shift the graph vertically.
b: an increase in the value would cause it to shift towards the left except when a<0 c: an increase in the value would cause it to move upwards.
When we complete the square, we get a(x+(b/2a)^2) + (4ac-b^2)/4a. Changing a changes the width of the parabola because lower a means that the inputs far away from -b/2a have lower outputs, whereas higher a means that the inputs far away from -b/2a have higher outputs.
The vertex of the parabola has coordinates (-b/2a, (4ac-b^2)/4a). Varying b chooses a vertex from -ax^2+c as seen in Why do the vertices of $f(x) = ax^2 + bx + c$, when fixing $a$ and $c$ but varying $b$, lie on $g(x) = -ax^2 + c$? Varying c simply shifts the graph up or down.