For a parabola that opens upward, is the vertex a minimum or a maximum point? What formula can you use to find the x-value of the vertex of $ax^2+bx+c$?

After you find the x-coordinate of the vertex, substitute that value back into $h(x)$ to get the corresponding y-value. That y-value is what the question is asking for.

In Desmos, type y = x^2 - 6x + 5 into an empty expression line. You will see a parabola opening upward.

On the graph, tap (or click) the lowest point of the parabola; Desmos will mark the vertex and show its coordinates. The y-coordinate of this point is the minimum value of $h(x)$.

Add a table for the expression and choose x-values around the vertex (for example, from $x=0$ to $x=6$). Look down the y-column and identify the smallest y-value; that is the minimum value of $h(x)$.

The function $h(x)=x^2-6x+5$ is a quadratic with leading coefficient $1>0$, so its graph is a parabola opening upward. For such a parabola, the minimum value of $h(x)$ occurs at the vertex of the parabola.

For a quadratic $ax^2+bx+c$, the x-coordinate of the vertex is given by

Here, $a=1$ and $b=-6$, so

So the vertex occurs at $x=3$.

Now plug $x=3$ into $h(x)$ to find the y-value of the vertex:

Because the parabola opens upward, this vertex y-value is the minimum value of $h(x)$, so the correct answer is $-4$.