Quadratic functions

The turning point of $y=a{x}^2$ sits on the origin. Add the term $2bx$ and a constant $c$, and that point leaves. Where does it land, and how many times do the arms still meet the x-axis?

The vertex stays at the origin

This picture keeps only $y=a{x}^2$, with $a\ne 0$. The sideways term and the constant stay out until the vertex is allowed to move.

The mark on the axis is the vertex, fixed at $(0,0)$. The other mark is the point at input $1$, so its height is $a$. While $a>0$ the arms open upward. Drag $a$ below zero and they open downward. The graph, with the vertex held at the origin, is a parabola. At $a=0$ both marks lie on the x-axis and the arms have flattened onto it, outside this family.

The vertex leaves the origin

Put $2bx$ and $c$ back. The opening is still set by $a$, and the vertex is free to move.

At $a=1$, $b=-3$, and $c=8$ the moving mark sits at $(3,-1)$. The faint curve is still $y=a{x}^2$, vertex at the origin. The live curve is $y=x^2-6x+8$.

The landing place in general is

$\left(-\frac{b}{a},-\frac{D}{a}\right),$

where

$D:=b^2-ac.$

That step is a transformation of functions: the vertex $(0,0)$ of $y=a{x}^2$ is sent to $\left(-b/a,-D/a\right)$, and every other point of $y=a{x}^2$ takes the same step. Completing the square names the curve that arrives there. The linear term is $2bx$, so

$y=a{x}^2+2bx+c=a{\left(x+\frac{b}{a}\right)}^2-\frac{D}{a}.$

How many times the arms meet the axis

Set the height to zero. For real coefficients and $a>0$,

$a{x}^2+2bx+c=0.$

On this board $a$ stays $1$ and $b$ stays $-3$, so $D=9-c$ and only $c$ moves. At $c=8$, $D=1$, and the marks are at $4$ and $2$. Drag $c$ to $9$: $D=0$, and the marks meet at $3$, where the arms touch the axis. Drag $c$ past $9$: $D<0$, and the real marks leave the figure.

For any real $b$ and $c$, and any $a>0$,

$x_{\pm }=\frac{-b\pm \sqrt{D}}{a}=\frac{-b\pm \sqrt{b^2-ac}}{a}.$

When $D>0$ there are two distinct real zeros, and $x_{+}$ is the one on the right. When $D=0$ there is one real zero, and the arms are tangent to the axis. When $D<0$ the zeros are the complex pair

$x_{\pm }=\frac{-b\pm \mathrm{i}\sqrt{-D}}{a}=\frac{-b\pm \mathrm{i}\sqrt{ac-b^2}}{a},$

where ${\mathrm{i}}^2=-1$. The real arms miss the axis.

The equation $x^2-6x+8=0$ is the board at $c=8$:

$x_{\pm }=3\pm \sqrt{3^2-8}=3\pm 1,$

so $x_{+}=4$ and $x_{-}=2$.

The equation $x^2-2x+1=0$ has

$x_{\pm }=1\pm \sqrt{1-1}=1.$

The equation $x^2+2x+2=0$ has

$x_{\pm }=-1\pm \sqrt{1-2}=-1\pm \mathrm{i}.$

A quadratic keeps the arms of $y=a{x}^2$ and only relocates the vertex. With $a>0$, the sign of $D$ decides two crossings, one touch, or a miss. The next question is what those arms do when the exponent on $x$ is free, which is the power function.