Vertex of a parabola with complex roots

Task

Joanne wants to graph a quadratic function whose roots are $5 \pm 2i$, and says: 

I know the graph is a parabola, and the roots tell me that my function does not cross the $x$-axis, but I'm not sure where to go next -- how do I use this information to help with my graph?

1. What can you deduce about the vertex of Joanne's parabola?
2. With the information provided, can you graph Joanne's function? Why?

IM Commentary

The goal of this task is to study the relationship between the roots of a quadratic function and its graph.  When the function has one real root, its graph is tangent to the $x$-axis. When it has two real roots, the vertex is halfway between the two roots. When there are no real roots, as happens in the case presented here, the complex roots still give us the $x$-coordinate of the vertex. In fact, even though the two roots are complex, it is still the case that the vertex is ''halfway'' between the two roots!  

The three cases for $x$-intercepts of the graph of a quadratic function can be understood if the function is expressed in terms of ''completing the square'' that is $$f(x) = a\left((x-h)^2 + k\right).$$ The minimum or maximum (depending on the sign of $a$) will always appear when $x = h$. When $k = 0$ the graph touches the $x$-axis at $x = h$. When $k \lt 0$ the function has two roots. Finally when $k \gt 0$ then the function is never equal to zero.  

Solution

1. Let $f$ be Joanne's quadratic function. We can write $f(x) = a\left(x^2 + bx + c\right)$ where $a$ is the leading coefficient. Since $f(5 + 2i) = 0$ we know that $(x - 5 - 2i)$ is a factor of $f$ and similarly $(x -5 + 2i)$ is a factor of $f$. This means that the quadratic function $$(x-5-2i)(x-5+2i) = x^2 -10x +29$$ divides $x^2 + bx + c$ (with no remainder). Since these both have leading coefficient 1 they must be equal and we conclude that $f(x) = a(x^2-10x+29)$. We can double check, applying the quadratic formula, that the roots of this quadratic function are indeed $5 \pm 2i$.  

   To find the vertex we can write $x^2 - 10x + 29 = (x-5)^2 +4$. This function takes a minimum value of 4 when $x = 5$ because $(x-5)^2$ is always non-negative and is 0 only when $x = 5$. If $a$ is positive, this means that $f(x)$ takes a minimum value of $4a$ when $x = 5$ and if $a$ is negative then $f(x)$ takes a maximum value of $4a$ when $x = 5$.  In summary, we know the $x$-coordinate of the vertex of this parabola must occur at $x=5$, but can't find its $y$-coordinate without further information.

2. No.  We know that $f$ is a parabola with a vertex at $x$-coordinate $5$, opening upward if $a$ is positive and downward if $a$ is negative. We cannot provide the exact graph, however, because we don't know the value of $a$. This value determines not only whether the graph opens up or down and the height of the vertex, but also how rapidly the function values increase (or decrease) as we move away from the vertex.