Question

factor, find the coordinates of the vertex of the related function, then graph.
$x^2 - 2x - 48 = 0$

factor.
$\square = 0$

Explanation:

Step1: Factor the quadratic trinomial

We need two numbers that multiply to $-48$ and add to $-2$. These numbers are $-8$ and $6$.
$x^2 - 2x - 48 = (x - 8)(x + 6)$
So the factored equation is:
$(x - 8)(x + 6) = 0$

Step2: Find vertex x-coordinate

For quadratic $ax^2+bx+c$, vertex x is $-\frac{b}{2a}$. Here $a=1$, $b=-2$.
$x = -\frac{-2}{2(1)} = 1$

Step3: Find vertex y-coordinate

Substitute $x=1$ into $f(x)=x^2-2x-48$.
$f(1) = (1)^2 - 2(1) - 48 = 1 - 2 - 48 = -49$

Answer:

Factored form: $(x - 8)(x + 6) = 0$
Vertex coordinates: $(1, -49)$
(Graph note: The parabola opens upward, crosses the x-axis at $x=8$ and $x=-6$, and has its lowest point at $(1, -49)$)