Track / Stage 4 / 18 of 340 #278 of 1964

Problem 278

AMC 12 late, AIME early

Algebra Difficulty 4.5 Multiple choice

4. The graph of the parabola $y=x^{2}+x+p(p \neq 0)$ intersects the $x$ -axis at one point with the $x$ -coordinate $p$ . Then, the coordinates of the vertex of the parabola are:

Pick one

Official solution

4.D.

From the problem, we get $p^{2}+p+p=0$ .
Solving, we get $p_{1}=-2, p_{2}=0$ (discard).
When $p=-2$ , the parabola is $y=x^{2}+x-2$ .
Therefore, the vertex coordinates are $\left(-\frac{1}{2},-\frac{9}{4}\right)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.