Mr Thompson, This is my homework! Please help.

Problem 1

There's not much to say about these. You'll just have to memorize the terms. Note with standard form that the exponents decrease as you move from left to right (2,1,0).

Problem 2

Other names for zeros are:

The term "root" is also applicable here.

Problem 3

We set each piece equal to 0 and solve for x. This is the zero product property

f(x) = (x-2)(x+4)

0 = (x-2)(x+4)

(x-2)(x+4) = 0

x-2 = 0 or x+4 = 0

x = 2 or x = -4

The two x intercepts are located at (2,0) and (-4,0)

Choice C

Problem 4

In the last problem, we found the two roots were x = 2 and x = -4

I'll let p = 2 and q = -4.

Average the roots to get (p+q)/2 = (2+(-4))/2 = -2/2 = -1

This is the x coordinate of the vertex. So h = -1

Plug this into the function to find k, which is the y coordinate of the vertex.

k = f(h)

k = f(-1)

k = (-1-2)(-1+4)

k = (-3)(3)

k = -9

The vertex is located at (h,k) = (-1, -9)

This is another way of saying the vertex is when x = -1 and y = -9

Choice C

all it means is function 3

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.