trig
posted by y912f on .
can you please show me in steps how to find f(1+a)
in this equation
f(x)=2x^2+4x+6
i know that you will plug in 1+a anywhere there is x. but after that i get confused.
also can you tell me how to find the domain and range of a relation
for ex. 2x+5y=10
thanks for your help!

f(x)=2x^2+4x+6
f(1+a) = 2(1+a)^2 + 4(1+a) + 6
= 2(1 + 2a + a^2) + 4 + 4a + 6
= 2 4a 2a^2 + 10 + 4a
= 2a^2 + 8
for 2x+5y=10 both the domain and range are the real numbers
You are looking at a slanted line, (slope=2/5), so you will find the graph for whatever x you use, and for whatever y you use.
This is true for all straight line graphs, which are neither vertical nor horizontal. 
ok,thanks alot.. what about this one : y=2x^25

this is a vertical parabola, meaning it opens either upwards or downwards
the domain of any vertical parabola is the set of real numbers
for the range:
if the parabola opens up, then y ≥ the y value of the vertex
if the parabola opens down, then y ≤ the y value of the vertex
in this case
Domain : x any real number
Range : y ≥ 5, y being real 
ok. so do u mean that to figure this out u always have to graph the relation?

no, you just have to find the vertex.
Here is a quick way to find the vertex:
if y = ax^2 + bx + c
then the x of the vertex is b/(2a)
sub that back into the function to find the y value. 
im sry im still kinda confused. can you show me once again please and explain everything you do thnks:
y=2x+3
y=4/square root of x^2+81
thnks alot!! 
btw, why are you calling this "trig" ?

its in the first chapter of my trig course. i know its review frm geometry and alg