Let me try,
P(x) = ax^2 + bx +c
So the vertex is at P'(x)=0
P'(x) = 2ax + b = 0
=> x = -b/2a
but given, b/2a = -1
so, x = 1
Therefore the axis of symmetry is at x=1
And let x-intercept be x1 and x2
So, (x1 + x2)/2 = x
=> x1 + x2 = 2*x = 2*1 = 2
=> x1 + x2 = 1 ------------------------ (1)
Also,
Given, x1 - x2 = 4 ------------------(2)
On solving eq (1) and eq(2) we get
x1=3 and x2= -1
so the equation is y = x^2 - (x1+x2)x + x1*x2
y = x^2 - 2x - 3
So, a = 1, b= -2 and c = -3
So, y position of vertex is given by -(b^2 - 4ac)/4a
=> y(vertex) = -(4+12)/4 = -4
So the coordinate of vertex is (1,-4)
Also y-intercept
y = 0^2- 2*0 -3 = -3
So,
Complete Solution:
x intercepts = -1 and 3
y intercept = -3
axis of symmetry: x=1
vertex position: (1,-4)
graph for y<=3 is in link below:
https://ibb.co/Js09K31