Graph the following quadratics and state its vertex and intercepts.
$y=x^2-2x-3$

• A. Vertex: $\(\left\)(-1,-4\(\right\))$; $x$-intercept: $\(\left\)(1,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• B. Vertex: $\(\left\)(1,-4\(\right\))$; $x$-intercept: $\(\left\)(3,0\(\right\)),\(\left\)(-1,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• C. Vertex: $\(\left\)(-1,-4\(\right\))$; $x$-intercept: $\(\left\)(1,0\(\right\)),\(\left\)(-3,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$
• D. Vertex: $\(\left\)(1,-4\(\right\))$; $x$-intercept: $\(\left\)(3,0\(\right\)),\(\left\)(-1,0\(\right\))$; $y$-intercept: $\(\left\)(0,-3\(\right\))$

Answer: (B)

Verified step by step guidance

1. Identify the quadratic function given: \(y = x^2 - 2x - 3\).
2. Find the vertex using the vertex formula for a parabola \(y = ax^2 + bx + c\), where the x-coordinate of the vertex is \(x = -\frac{b}{2a}\). Here, \(a=1\) and \(b=-2\), so calculate \(x = -\frac{-2}{2 \times 1}\).
3. Substitute the x-coordinate of the vertex back into the quadratic equation to find the y-coordinate of the vertex: \(y = (x)^2 - 2(x) - 3\).
4. Find the x-intercepts by setting \(y=0\) and solving the quadratic equation \$0 = x^2 - 2x - 3$. Factor the quadratic or use the quadratic formula to find the roots.
5. Find the y-intercept by setting \(x=0\) in the equation, which gives \(y = 0^2 - 2(0) - 3\).