Compare $f(x) = 3(x - 2)^2 + 4$ with $y = a(x - h)^2 + k$. What value of $h$ makes $(x - h)$ match the $(x - 2)$ part?

In vertex form, which part of the equation represents how far the graph is moved up or down? What is that value in this function?

Once you know $h$ and $k$, remember that the vertex is written as the point $(h, k)$. Write that coordinate pair explicitly.

In Desmos, type y = 3(x - 2)^2 + 4 into an expression line to graph the parabola.

Zoom or move the graph as needed, then click or tap on the lowest point of the parabola (its minimum). Desmos will display a point there with its coordinates; those coordinates are the vertex of the graph.

Compare the vertex coordinates shown in Desmos with the listed options and select the matching coordinate pair.

A parabola written as $y = a(x - h)^2 + k$ is in vertex form. In this form, the vertex of the parabola is the point $(h, k)$.

Compare $f(x) = 3(x - 2)^2 + 4$ with $y = a(x - h)^2 + k$.

• The coefficient $a$ is $3$ (this affects the width and direction, not the vertex location).
• Inside the parentheses, we have $(x - 2)$, which matches $(x - h)$, so $h = 2$.

Be careful: because the form is $(x - h)$, if you see $x - 2$, then $h$ is $+2$, not $-2$.

The constant term outside the square in vertex form is $k$.

In $f(x) = 3(x - 2)^2 + 4$, the constant outside is $+4$, so $k = 4$.

This means the graph is shifted up 4 units from the basic parabola $y = x^2$.

The vertex of a parabola in vertex form $y = a(x - h)^2 + k$ is $(h, k)$.

Here, $h = 2$ and $k = 4$, so the vertex is $(2, 4)$, which corresponds to choice D.