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

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 $$y = x^2$$?

• +1 for beautiful graphs and your efforts too!! – StammeringMathematician Sep 27 '18 at 4:18
• @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator – Slecker Sep 27 '18 at 4:22
• +1 from me as well. This attitude should be highly encouraged here on MSE. – Ahmad Bazzi Sep 27 '18 at 4:34
• Slecker. Beautiful +. – Peter Szilas Sep 27 '18 at 9:17

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.

• 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 ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? – Slecker Sep 27 '18 at 4:46
• @Slecker I'm not familiar with the term "vertex form," but I would say that you are correct. – saulspatz Sep 27 '18 at 5:04

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?

• 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. – Slecker Sep 27 '18 at 15:22
• 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. – Peter Szilas Sep 27 '18 at 17:56
• Ah ok thanks for the clarification! – Slecker Sep 27 '18 at 18:26
• Slecker. If anything else, just say so:) – Peter Szilas Sep 27 '18 at 18:35