3 years ago

I need help with my algebra II!!

1 answer:

3 0

1st:
f(x) = x^2.
f(x +1): Anywhere we see x we replace it with (x+1)

f(x+1) = (x+1)^2 , Expand
(x+1)^2 = (x+1)(x+1) = x^2 + 2x + 1= C.

2nd:
Noting general equation, y = mx + c
m = slope, c = vertical intercept.
From the graph, slope = (2,0) and (0, -4)
= (-4 -0) / (0 - 2) = -4/-2 = 2.
Vertical intercept, c = -4.
y = mx + c , y = 2x + (-4), y = 2x - 4 = D.

3rd:
f(x) = x^2 - x, f(-4). we would substitute x = -4.

f(-4) = (-4)^2 - (-4) = 16 + 4 = 20 = D.
Cheers.

You might be interested in

I need help please take a look at this

Reika [66]

Answer:

$y = x-1$

$-x+\dfrac32=-y\\\implies y=x-\dfrac32$

The equations have the same slope but different y-intercepts. This means the lines are parallel and there is no solution (since they do not intersect).