Question

find the vertex of the parabola y=2x^2-10x+1.

Answer 1

Given parabola y = 2x^2-10x+1

Compare it standard parabola form y = ax^2+bx+c

a = 2, b = -10, c = 1.

Axis of symmetry x = -b/2a = 10/4

x = 5/2

To find vertex substitute x value in given equation.

y = 2(5/2)^2-10(5/2)+1

y = 2(25/4)-25+1

y = 25/2-24

y = -23/2

Vertex of parabola = (5/2,-23/2)