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^2-5
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=4/square root of x^2+81
btw, why are you calling this "trig" ?
its in the first chapter of my trig course. i know its review frm geometry and alg