## The problem

how did you graph it?

## Answer provided by our tutors

The graph of a quadratic function (such is also "y = - (1/2)x^2") is always U-shaped curve called a parabola.

The orientation of a parabola is that it either opens up or opens down, since there is (-1/2) in front of x^2, the **U-shaped curve opens down**.

The vertex (h, k) is the highest point on the graph "y = - (1/2)x^2" and it is calculated by the formula

h = -b/2a where a = (-1/2), b = 0

We plug the values for "a" and "b" and find:

h = -0/(2*(1/2))

h = 0

"k" is calculated by plugging "x = 0" into "y = - (1/2)x^2" so "k=0"

The **vertex is (h,k) = (0, 0)**The axis of symmetry is the vertical line that goes through the vertex, dividing the parabola into two equal parts. If h

is the x-coordinate of the vertex, then the equation for the

**axis of symmetry is x=0.**

We also need a point from the graph, so we can plug "x = 2" and calculate "y":

y = (-1/2)2^2

y = -2

Same for "x = -2" we find that "y = -2"

Now we have 2 points of the graph: (2, -2) and (-2, -2) and we can draw: