Given the function ƒ(x) = x^2 - 4x - 5 1. identify the zeros using factorization. 2. draw a graph of the function. its vertex is at (2, -9).

F(x) = x² - 4x -5, factored is simply f(x) = (x - 5)(x + 1).

if you zero out f(x), namely 0=(x-5)(x+1), that simply gives you the zeros of 5 and -1.

now, from -105, notice from -1 to 5, there are 6 units, and half of 6 is 3, so, the half-way is 3 units away from either zero.

so we can get to the half-way point by -1 + 3, or 2, x = 2, that's where the vertex is at, but you already knew that, and of course the y-coordinate is at f(2) = (2)² - 4(2) - 5, which is -9, so the vertex is indeed at (2, -9).

now to draw it, well, surely you can, simply use the zeros location, -1 and 5, and draw a bowl between them, with the bottom of the bowl at 2, -9.

Something that is true about that is that it has numbers