vertex, axis of symmetry and x-intercepts?

Identify the vertex, axis of symmetry and x-intercepts. f(x)= -x^2-4x+1

thanks for all the help

2 Answers

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  • DWRead
    Lv 7
    7 years ago
    Best Answer

    y = -x² - 4x + 1

    The coefficient of the x² term is negative, so the parabola opens downwards.

    The vertex-form equation of a down-opening parabola has the form

     y = a(x - h)² + k

    where

     a < 0

     (h, k) is the vertex

     x = h is the axis of symmetry

    Convert y = -x² - 4x + 1 to vertex form:

     factor out leading coefficient

     y = -1(x² + 4x) + 1

     complete the square

     y = -1(x² + 4x + 2²) + 2² + 1

     y = -1(x + 2)² + 5

    Now you have all the information you need.

  • 7 years ago

    f(x)=-(x+2)^2+5

    (-2,5) , x=-2 , -2 +/- sqrt(5)

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