"<span>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.
</span><span>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).</span>
