Consider the parabola $\rho$ given be the equation $y = ax^2 + bx + c$. Recall that varying $a$ in this equation stretches/squashes $\rho$ and that varying $c$ shifts $\rho$ vertically. The change in the parabola $\rho$ as you vary the value of $b$ is a bit harder to describe, but as you vary $b$, the vertex of $\rho$ traces out a familiar curve. What is this curve? What it the equation of this curve?
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First let's write down the general form for the vertex of $\rho$. Through any of various means we can find that the $x$-coordinate of the vertex is $-\frac{b}{2a}$. Substituting this value into the equation for $\rho$ we get that the vertex is
$$\left(-\frac{b}{2a},\;\; c-\frac{b^2}{4a}\right).$$
Since the $x$-coordinate of the vertex is linear in $b$ and the $y$-coordinate is quadratic in $b$ (has a $b^2$), we can conclude that the vertex of $\rho$ traces out a parabola. We can reparametrize this curve to get an explicit equation to describe it. Letting $t = -\frac{b}{2a}$ we can make the substitution $b = -2at$ into the $y$-coordinate, giving us the parametric curve defined by
$$ \left(t,\;\; c-\frac{(-2at)^2}{4a}\right) \;\;=\;\; \left(t,\;\; c-at^2\right) .$$
To state this explicitly, the curve traced by the vertex of $\rho$ is given by the equation $y = c - ax^2$.
answered Apr 4 '16 at 3:51
