I have a simple graph question, but I can see a few potential answers.
The graph of $y = x^2 + px + q$ has a turning point at $(2,5)$. Find the values of $p$ and $q$.
Can someone explain why the correct answer is $p = -4 \text{ and } q = 9$?
The answer I got was $p = 0, q = 1$ and another answer I got was $p = -1, q = 3$.
On
We are told that the turning point is at $(2,5)$.
We are given the leading coefficient is $1$, the quadratic curve is $$(x-2)^2+5=x^2-4x+9$$
$p$ and $q$ are unique.
In general if the turning point is $(x_0, y_0)$ with the leading coefficient being $1$, the quadratic curve is
$$(x-x_0)^2+y_0=x^2+(-2x_0)x + (x_0^2+y_0)$$
The $p$ that you proposed doesn't satisfy the formula.
$y=x^2+px+q$
$5=4+2p+q$
So $p$ and $q$ where $2p+q=1$ will have $(2,5)$ on the curve. It isn't a turning point though unless $2x+p=0 $ and $x=2 \implies p=-4$
Not a vertex: