What is the vertex of the parabola given by the equation y x^2+ 4?

Format of Quadratic Equation:  y = ax2 + bx + c

Given Quadratic Equation:  y = 2x2 - 3x + 3

Coefficient Variable Values:  a = 2 and b = -3 and c = 3

Axis of Symmetry:  x = -b/2a = -(-3)/2(2) so answer is x = 3/4

Vertex:  x value is axis of symmetry (3/4) and y value is calculated substituting 3/4 for x in original equation:  y = 2(3/4)2 - 3(3/4) + 3 = 2(9/16) - 9/4 + 3 = 9/8 - 9/4 + 3 = 9/8 - 18/8 + 24/8 = 15/8,
so answer is (3/4,15/8)

x intercepts (solve using quadratic formula):  x = (-b plus or minus sqrt(b2 - 4ac)/2a, so plugging in coefficient values for a and b and c, we get x = [-(-3) plus or minus sqrt((-3)2 - 4(2)(-3)]/2(2), which results in x = (3 + sqrt(33))/4 or (3 - sqrt(33))/4 and answers to nearest tenth are x = (3 + 5.7) / 4 = 2.2
or x = (3 - 5.7) / 4 = -0.7

y intercept is calculated by substituting zero for x into original equation:  y = 2x2 - 3x + 3, so y-intercept is 3.

Domain is range is from calculated negative x intercept (-0.7) to calculated positive x intercept (2.2) and is written as (-0.7,2.2)

Range is from calculated y-intercept to positive infinity, since parabola opens up due to positive x2 coefficient value, so range is written as (3,positive infinity).  Note:  infinity symbol is sideways 8.