SOLUTION: Two function f and g are defined by the rules f(x) = (x^2-6x) and g(x) = ( x+4). Represent the follwing in a graph

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Two function f and g are defined by the rules f(x) = (x^2-6x) and g(x) = ( x+4). Represent the follwing in a graph      Log On


   



Question 672614: Two function f and g are defined by the rules f(x) = (x^2-6x) and g(x) = ( x+4). Represent the follwing in a graph
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+%28x%5E2-6x%29 and g%28x%29+=+%28+x%2B4%29
f%28x%29+=+%28x%5E2-6x%29..=>...f%28x%29+=+x%28x-6%29; so, roots are

x%28x-6%29=0
if x=0, then x=0........one point is (0,0)
if x-6=0, then x=6.....second point is (6,0)
find few more points and make a table
f%28x%29+=+%28x%5E2-6x%29..let x=1
f%281%29+=+%281%5E2-6%2A1%29.....=> f%281%29+=+1-6...=> f%281%29+=+-5...third point is (1,-5)
f%28x%29+=+%28x%5E2-6x%29..let x=-1
f%28-1%29+=+%28%28-1%29%5E2-6%2A%28-1%291%29.....=> f%28-1%29+=+1%2B6...=> f%28-1%29+=+7...third point is (-1,7)

table
x|f%28x%29
0|0
6|0
1|-5
-1|7
plot these points and draw a graph of a parabola

for g%28x%29+=+%28+x%2B4%29 root is
if +%28+x%2B4%29=0, then x=-4
this is linear function and we will need only two points to graph a strait line

x|f%28x%29
0|4
-4|0


+graph%28+600%2C+600%2C+-10%2C+20%2C+-10%2C+20%2C+x%5E2-6x%2C+x%2B4%29+