Once you have $y = a(x - 3)^2 - 4$, how can you use the fact that the graph passes through $(0, 5)$ to solve for $a$?

After you find $a$, expand your expression so it looks like $y = x^2 + bx + c$. Then see which answer choice has exactly the same $x^2$, $x$, and constant terms.

In Desmos, enter each option on its own line: y = x^2 + 6x + 5, y = (x+3)^2 - 4, y = (x-3)^2 + 4, and y = x^2 - 6x + 5. You should see four different parabolas.

For each graph, click on the lowest point (the vertex) of the parabola. Desmos will display its coordinates. Identify which graph has vertex $(3, -4)$.

For that graph, either click on $x=0$ or add a table and check the row where $x=0$. Confirm that the corresponding $y$-value is 5. The equation whose graph has vertex $(3, -4)$ and passes through $(0, 5)$ is the one you should choose.

When you know a quadratic’s vertex, it is easiest to start with vertex form.

Vertex form is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

Here the vertex is $(3, -4)$, so substitute $h = 3$ and $k = -4$ to get:

The value of $a$ is still unknown.

The graph passes through $(0, 5)$, so when $x = 0$, $y$ must be $5$.

Substitute $x = 0$ and $y = 5$ into $y = a(x - 3)^2 - 4$:

This simplifies to $5 = 9a - 4$, so $9 = 9a$ and $a = 1$.

So the equation becomes $y = (x - 3)^2 - 4$.

Now expand $y = (x - 3)^2 - 4$ to compare with the answer choices.

$(x - 3)^2 = x^2 - 6x + 9$, so

y = $x^2 - 6x + 9 - 4 = x^2 - 6x + 5$.

So the quadratic is $y = x^2 - 6x + 5$, which corresponds to choice D.