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Algebra

posted by on .

f(x)=1-x^2.
Looking for the vertex and an equation of the line of symmetry.

f(x)=-x^2+5x+36 Looking for the x, and Y coordinates.

I seem to be entering the answers backwards, cannot understand.

Please help.

  • Algebra - ,

    shortcut to find vertex of
    y = ax^2 + bx + c
    the x of the vertex is -b/(2a)
    in your case, the x of the vertex = 0/-2 = 0
    sub back in
    f(x) = 1
    vertex is (0,1)
    axis of symmetry is x = 0

    2nd:

    you must mean the x and y -intercepts
    let y or f(x) = 0
    x^2 - 5x - 36 = 0
    (x-9)(x+4) = 0
    x = 9 or x = -4

    for y-intercept let x = 0
    then y = f(0) = 36

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