Question 542528
{{{y = 2x^2-6x+3}}}
You have correctly found the x-coordinate of the vertex:
{{{x = 3/2}}} and you have made a start on finding the corresponding y-coordinate by substituting {{{x = 3/2}}} into the given quadratic equation to solve for y.
{{{y = 2(3/2)^2-6(3/2)+3}}} Simplify.
{{{y = 2(9/4)-6(3/2)+3}}}
{{{y = 9/2-18/2+6/2}}}
{{{y = -3/2}}}
The coordinates of the vertex are: (3/2, -3/2)
The vertex is a minimum (the parabola opens upward) which is indicated by the positive coefficient of the {{{x^2}}} term.
The axis of symmetry is give by {{{x = 3/2}}} which is the equation of the vertical line passing through the point (3/2, -3/2). (the vertex.)
The range consists of all of the valid y-values.
Since the y=coordinate of the vertex (a minimum) is {{{y = -3/2}}} it is clear that the range is {{{-3/2}}} and above.
So the range can be written as:
{{{y>=-3/2}}}
You can see this from the graph:
{{{graph(400,400,-5,5,-3,10,2x^2-6x+3)}}}