The problem asks us to analyze the quadratic function $y = x^2 - 2x$ by finding its y-intercept, x-intercepts, and vertex.

1. Problem Description

The problem asks us to analyze the quadratic function $y = x^2 - 2x$ by finding its y-intercept, x-intercepts, and vertex.

2. Solution Steps

To find the y-intercept, we set $x = 0$ in the equation:

$y = (0)^2 - 2(0) = 0$

So the y-intercept is $(0, 0)$.

To find the x-intercepts, we set $y = 0$ and solve for $x$:

$x^2 - 2x = 0$

$x(x - 2) = 0$

Thus, $x = 0$ or $x = 2$.

The x-intercepts are $(0, 0)$ and $(2, 0)$.

To find the vertex, we can complete the square to rewrite the quadratic function in vertex form, which is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

$y = x^2 - 2x$

$y = (x^2 - 2x + 1) - 1$

$y = (x - 1)^2 - 1$

Thus, the vertex is $(1, -1)$.

Alternatively, the x-coordinate of the vertex of a quadratic $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. In this case, $a = 1$ and $b = -2$, so $x = -\frac{-2}{2(1)} = \frac{2}{2} = 1$. To find the y-coordinate, we substitute $x = 1$ into the equation:

$y = (1)^2 - 2(1) = 1 - 2 = -1$

Thus, the vertex is $(1, -1)$.

3. Final Answer

y-intercept: (0, 0)

x-intercepts: (0, 0), (2, 0)

Vertex: (1, -1)