MathQuestion again. Help!?

Question

Let f be a function such that f(-2)=5,f(0) = -2, and f(1) = 0. Give the coordinates of three points on the graph of A. y=f(x+3) b. y=f(x)+1

Answer 1

The key to finding the co-ordinates of 3 points on the graph of y is to choose the values of x so that y is formed by f(-2), f(0) and f(1) since those f functions' values are what you know.

For example, y = f(x+3), let x1 = -5, y1 = f(-5+3) = f(-2) = 5.

-> 1 point on the graph of y=f(x+3) is (-5,5)

Answer 2

Looks like a parabola of form ax^2 +bx +c = 0
Since f(0)= -2 , c = -2, giving f(x) = ax^2 +bx - 2= 0
Since f(-2) = 5, we get 4a -2b = 7
Since f(1) = 0, we get a+b =2
Solving these two equations simultaneously gives:
a=11/5 and b = - 1/5
Therefore f(x) = (11/5)x^2 -(1/5)x - 2 =0

f(x+3)=(11/5) (x+3)^2-(1/5)(x+3) -2
=(11/5)(x^2 +6x +9) -(1/5)x -3/5 -2
=(11/5)x^2 +(66/5)x +(11/5)9 -(1/5)x -2 3/5
= (11/5)x^2 +13x +86/5

f(x)+1= (11/5)x^2 - (1/5)x -1

Answer 3

let y=mx+b
5=m(-2)+b
=>-2m+b=5
f(1)=0=m+b
=>m+b=0
subtracting eqn.2 from eqn.1
-3m=5
m=-5/3
substituting
b=5/3
so the equation
f(x)=-5/3 x+5/3
f(x+3)
=-(53)(x+3)+5/3
f(x)+1=[-5/3x+5/3]+1