How do you find the axis of symmetry when the vertex are (4, 0)?

Question

I had to find the coordinates of the vertex and the equation of the axis of symmetry for the parabola given the by the equation h(x)=-1/2x to the 2nd power +4x-8 and the answer I came up with the answer (4,0) as my x and y coordinate.

Answer 1

y =-1/2 X² + 4X - 8=-1/2(x²-8x+16)=-1/2(x-4)²

-2y=(x-4)² or -4(1/2)y=(x-4)².  Which gives that the focal length is 1/2, that the vertex is (4,0),

and that the focus is (4,-1/2,) and the directrix is the line y=1/2.  This all comes from the form for the parabola

(x-h)²=4p(y-k).  Vertex (h,k) focal length p.  The axis of symmetry is the line x=h, or in our case x=4.

Answer 2

h( x ) =-1/2 X ^2 + 4X - 8 

 

  Always quadratic is represented in general form by:

 

    a X^2 + b X + c

 

   ( a ,b, c ) are always referred to as the coefficients of ( 2nd power, first power, constant) in a polynomial of 2nd degree, a Trinomial.

 

    Coordinate of  Vertex is always : ( -b/2a , ( b^2 - 4ac) / 4a^2)

 

    Axis of Symmetry  always : ( X = -b/2a) , a vertical line drawn from the vertex.

   

   In this case

 

     f( X ) = -1/2 X^2 + 4X - 8

 

             a = -1/2    b = 4   c = -8

 

     axis of Symmetry :  X = -b/ 2a = - 4/( 2(-1/2)) = 4 

 

                                  X = 4 is the axis of Symmetry:

Answer 3

Alicia

The vertex is on the axis of symmetry  at x=-b/2a=-4/-1=4 sub this in the original quadratic and you get h(4)=0 so the coordinates are (4,0)

 

Jim