Consider the function $q(x)=ax^2+bx+c$, where $a \ne 0$ (so that it's truly quadratic, and not linear).

We want to consider several features of $q$, including

• shape (a parabola)
• the focus
• the roots

Equivalent expressions for $q(x)$ are

• $q(x)=ax^2+bx+c$
• $q(x)=a(x-m)^2+h$
• $q(x)=a(x-r_1)(x-r_2)$

The values of $m$ and $h$, and the roots $r_1$ and $r_2$, are given using basic algebra:

• The vertex of the parabola is at $(m,h)=(\frac{-b}{2a},c-\frac{b^2}{4a})$
• The roots of the parabola (where it intercepts the $x$-axis) are at $\{r_1,r_2\}=\{\frac{-b \pm \sqrt{b^2-4ac}}{2a}\}$
• The focal point of the parabola is at a height where the slope of the tangent line is 1, at $(m,p)=(\frac{-b}{2a},h+\frac{1}{4a})$

$a \gt 0$: A bowl (with a "double root"):

xxxxxxxxxx

a=1

b=-2

c=1

h=c-b^2/(4*a)

p=1/(4*a)+h

f(x) = a*x^2+b*x+c

g(x) = h

e(x) = p

p1=plot(f, (x, -2, 4), color='green', thickness=3)

p2=plot(g, (x, -2, 4), color='blue', thickness=1)

p3=plot(e, (x, -2, 4), color='red', thickness=1)

p1+p2+p3

$a \lt 0$: An umbrella, with symmetric roots:

xxxxxxxxxx

a=-1

b=0

c=1

h=c-b^2/(4*a)

p=1/(4*a)+h

f(x) = a*x^2+b*x+c

g(x) = h

e(x) = p

p1=plot(f, (x, -2, 2), color='green', thickness=3)

p2=plot(g, (x, -2, 2), color='blue', thickness=1)

p3=plot(e, (x, -2, 2), color='red', thickness=1)

p1+p2+p3