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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.

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

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

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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:

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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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    $\begingroup$ +1 for beautiful graphs and your efforts too!! $\endgroup$ Sep 27, 2018 at 4:18
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    $\begingroup$ @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator $\endgroup$
    – Slecker
    Sep 27, 2018 at 4:22
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    $\begingroup$ +1 from me as well. This attitude should be highly encouraged here on MSE. $\endgroup$ Sep 27, 2018 at 4:34
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    $\begingroup$ Slecker. Beautiful +. $\endgroup$ Sep 27, 2018 at 9:17
  • $\begingroup$ Note that no matter what the value of $b$ is, the graph of $y = ax^2 + bx + c$ must always pass through the point $(0,c).$ That's why your "$bx$" curves all pass through one point. It also implies you can't just shift the curve sideways, because it would then lose contact with that point. $\endgroup$
    – David K
    Jun 4, 2022 at 15:27

2 Answers 2

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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.

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  • $\begingroup$ It took me a while to realize that you transformed it into vertex form. So would the reason that $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertex, since the vertex coordinates are ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? $\endgroup$
    – Slecker
    Sep 27, 2018 at 4:46
  • $\begingroup$ @Slecker I'm not familiar with the term "vertex form," but I would say that you are correct. $\endgroup$
    – saulspatz
    Sep 27, 2018 at 5:04
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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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  • $\begingroup$ I'm having a hard time understanding what you mean by "Revert to original $x$, $y$ coordinates" and where the subsequent equation, $y-y_0 = a(x-x_0)^2$, came from. I think once I understand that I can interpret the rest of your answer. $\endgroup$
    – Slecker
    Sep 27, 2018 at 15:22
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    $\begingroup$ Slecker.Draw two coordinate systems, x,y and another one ,call it x',y'.Say, you put the origin of the x',y' system at x_0=3, y_0=4.x'y' system has its origin at (x_0,y_0)=(3,4), ok?. put a normal parabola y'=ax'^2 in the x',y' system.x'=1; y'=a; everything in x'y'.Take any x' coordinate, say x'=7, what is the x value in the original system: x= 7+ 3= x' +x_0 ok? Likewise y= y'+y_0. Solve for x' and y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2, now you are back in the original system.Your b =-2ax_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. $\endgroup$ Sep 27, 2018 at 17:56
  • $\begingroup$ Ah ok thanks for the clarification! $\endgroup$
    – Slecker
    Sep 27, 2018 at 18:26
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    $\begingroup$ Slecker. If anything else, just say so:) $\endgroup$ Sep 27, 2018 at 18:35

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