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I noticed a small thing while playing with the graph of quadratic. $$ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^2 + c - a\left(\frac{b}{2a}\right)^2$$

Clearly $b,c$ only determine how the vertex of the graph changes, not the shape of the graph; that is, as $b,c$ are varied, the graph just translates without changing its shape.

This means adding a linear function $bx+c$ to a quadratic doesn't change its shape! This makes sense from above crude manipulation of the equation but I'm wondering if there is a more satisfying way to see this, perhaps geometry/calculus?

user
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across
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    This will arise with any curve of equation $y=f(x)$, when you turn it to $y=f(x-X)+Y$: the graph translates by $(X,Y)$. When $f$ is $ax^2$, the nonlinear term remains $ax^2$. –  Sep 14 '21 at 16:36

1 Answers1

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Given

$$y=ax^2+bx+c$$

by a translation $y=Y+A$ and $x=X+B$ we obtain

$$Y+A=a(X+B)^2+b(X+B)+c$$

$$Y=aX^2+\overbrace{(2aB+b)}^{\beta}X+\overbrace{aB^2+bB+c-A}^{\gamma}$$

$$Y=aX^2+\beta X+\gamma$$

which has the same shape.

We can also go backward and show that adding a linear term corresponds to a translation, therefore shape doesn't change.

Refer also to the related

user
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  • If I understand correctly, $B, A$ can be uniquely determined from the given quadratic: $B = \frac{\beta - b}{2a}, A = aB^2 + bB+c-\gamma$. I don't see right away why this fails for a cubic like $ax^3+bx+c$ ? I know the shape changes as $b,c$ are varied for cubic, but I don't know why.. something to think later(for me). Thank you so much for the awesome answer:) – across Sep 14 '21 at 17:12
  • @across For a cubic the $ax^3$ term gives $a(X+B)^3=aX^3+3aBX^2+\ldots$. Therefore in general we also need to add quadratic terms to have the same shape. You are welcome! Bye – user Sep 14 '21 at 17:16