Question

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).

Answer 1

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 ----- -1------0---------------------------------5, 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.

Answer 2

Answer:

(x+1)(x-5)

For graph see image

Step-by-step explanation:

In order to solve this you first think of two numbers that multiplied by one another would give as a result -5 and added up together would result in -4:

-5*1= -5 ----- -5+1= -4

5*-1= -5 ------ 5-1= 4

So from the options we know that the only one that works is the pair of -5, 1, so those are our zeros for factorization:

x^2-4x-5=0

(x-5)(x+1)=0