Y= -x^2-3x. I need to find the vertex, axis of symmetry y and x intercepts

Question

someone please help me get the answers to this problem please and thanks 

Answer 1

For a quadratic function in standard form (y=ax² + bx +c), the vertex is the point (h,k), and the axis of symmetry is x=h. 

Here are the formulas for h and k:
h=-b/(2a)

k=c-ah²

 

For your problem, the function is y=-x² - 3x, a=-1, b=-3, and c=0. Using those formulas you can figure it out.

Answer 2

If y = Ax²+Bx+C, then the x-coordinate of the vertex is -B/(2A).

 

So, for the given function, the x-coordinate of the vertex is                  x = -3/2

 

                              y-coordinate of vertex = -(-3/2)²-3(-3/2)

 

                                                             = -9/4 +9/2 = 9/4

 

         vertex:  (-3/2, 9/4)

 

         axis of symmetry:  x = -3/2

 

To find the x-intercepts, solve the equation -x²-3x = 0

 

                                                          -x(x+3) = 0

 

                                                           x = 0 or -3

 

         x-intercepts:  (0,0) and (-3,0)