Question

F(x)=-1/2x^2+4vertex:min or max:axis of symmetry:y-intercept:domain:range:

Answer 1

"F(x)=-1/2x^2+4           Important:  enclose -1/2 inside parentheses.
vertex:                          For quadratic y=ax^2+bx+c, x value at vertex is -b/2a.
min or max:            Corresponds to vertex x value; must also find y value
axis of symmetry:    Same as x=value of vertex
y-intercept:             Set x=0 and solve for y
domain:           Domain is "all real numbers" when f(x) is a polynomial.
range:"          Here, the graph opens down (because (-1/2) is negative), so the
                      range begins at -infinity and extends up to and including the y-
                       value at the vertex.

F(x)= (-1/2)x^2+4

vertex:  Since a = -1/2 and b = 0, x-value at vertex is -b/(2a), or 
        -(0) / (2[-1/2]) = 0.  Graph is symmetric about the y-axis.
         The function is F(x) = (-1/2)x^2 +4.  At x = 0, y=4.
          Thus, the vertex is at (0,4).

maximum occurs at the vertex, and is (0,4).  Max value of function is 4.

axis of symmetry:  see "vertex," above.  x = 0 is that axis.

domain:  since this is a polynomial function, the domain is the set of all real numbers.

range:  (-infinity, 4]      This includes y =4 and all smaller y values.

y-intercept:  Set x=0 and find y.  As before, y=4.  y-intercept is (0,4).