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 $y = x^2$?
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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?
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