How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given #y=1/2x^2+x-2#?

1 Answer
Nov 30, 2017

The graph (the Parabola) Opens Up, as the coefficient of #x^2# term is greater than ZERO.

Vertex #= (-1, -2.5)# and Axis of Symmetry #x = -1# Graph attached for a visual proof.enter image source here

Explanation:

Standard Form of a Quadratic Equation is given by
#ax^2 + bx + c = 0#

If #a > 0# then the Parabola will "Open Up". If #a < 0# then the parabola will "Open Down".

Since in our Quadratic Equation, the coefficient of the #x^2# term is found to be #>0#, the Parabola "Opens Upward".

Also observe that, if the Parabola "Opens Upward" then our Vertex will be a "Minimum" of the Quadratic Function.

We are given the Quadratic Function #y = (1/2)x^2 + x - 2#

The expression #(-b/(2a))# will give us the x-coordinate value of the Vertex.

To obtain the y-coordinate value of the "Vertex", substitute the value obtained from using the expression #(-b/(2a))# in our original equation.

Hence, #y = (1/2)(-1)^2 + (-1) -2# will help us to get the y-coordinate value of the Vertex.

After simplification, we get #y = -2.5#

Hence our Vertex# = (-1, -2.5)#

And our Axis of Symmetry #x = -1#

Please refer to the attached graph for a visual proof of our solution.

I hope this explanation is helpful.