How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$?

Question

Every parabola represented by the equation $y = ax^2 + bx + c$ can be obtained by stretching and translating the graph of $y = x^2$.

Therefore:

The sign of the leading coefficient, $-a$ or $a$, determines if the parabola opens up or down i.e.

The leading coefficient, $a$, also determines the amount of vertical stretch or compression of $y = x^2$ i.e.

The constant term, $c$, determines the vertical translation of $y = x^2$ i.e.

Now for $bx$. Initially, I thought it would determine the amount of horizontal translation since the constant term, $c$, already accounted for the vertical translation, but when I plugged in some quadratics the graph of $y = x^2$ translated both horizontally and vertically. Here are the graphs:

Seeing as the middle term, $bx$, does more than just horizontally translate, how do you describe its effect on $y=x^2$? Would it be accurate to say that it both horizontally and vertically translates the graph of $y = x^2$?

Answer 1

Yes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\left(x+\frac32\right)^2-\frac94$$

Compare that to your graph of $y=x^2+3x$. Of course, if the coefficient of the quadratic term is not $1$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square.

Answer 2

Look at $2$ Cartesian coordinate systems $X,Y$ and $X',Y'$.

Origin of $X',Y$' is located at $(x_0,y_0)$, $X'$-axis parallel $X$-axis , $Y'$-axis parallel $Y$-axis(a translation),i.e.

$x= x_0+x'$; $y= y_0+ y'$.

Set up your normal parabola in the $X',Y'$ coordinate system.

$y'=ax'^2$, vertex at $(0',0')$.

Revert to original $x,y$ coordinates .

$y-y_0= a(x-x_0)^2$ ;

$y=ax^2 -2(ax_0)x +ax_0^2$.

Compare with $y =ax^2+bc +c$:

$b=-2ax_0$.

Can you interpret?