Question 745985
First solve the equation: {{{y=-2x^2 + 4x+1}}}
Since the squared term is x, equate the system in x, thus,
{{{2x^2-4x+y-1=0}}}, completing the square
 
{{{2x^2-4x+2+y-1=2}}}, add two both side
{{{2(x^2-2x+1)=-y+3}}},
{{{(x^2-1)^2=(-1/2)(y-3)}}}
The parabola is facing downward,
*[invoke quadratic "x", -2, 4, 1 ]
.
.
so, the vertex:(1,3)
x-intercept: (-2.22,0),(0.22,0)
y-intercept: (0,1)