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

Question

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?

Answer 1

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

• Related to parabola