# Graph of quadratic $f(x)=ax^2+bx+c$ when $a$ is fixed and $b,c$ are varied

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?

• 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 $ax^2$.
– user958916
Commented Sep 14, 2021 at 16:36

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

• 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:) Commented Sep 14, 2021 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
Commented Sep 14, 2021 at 17:16