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can you please show me in steps how to find f(1+a)
in this equation
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!

  • trig -

    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.

  • trig -

    ok,thanks alot.. what about this one : y=2x^2-5

  • trig -

    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

  • trig -

    ok. so do u mean that to figure this out u always have to graph the relation?

  • trig -

    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.

  • trig -

    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

    thnks alot!!

  • trig -

    btw, why are you calling this "trig" ?

  • trig -

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

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