I was wondering if anyone can tell me if I got the right ans

I was wondering if anyone can tell me if I got the right answer for this f(x) -3x^2+3x-2

opens down, maximum, vertex is (1/2, -5/4) domain: no real soultion, range: -5/4, axis of symmetry is to the y- axis, x--intercepts: none, y-intercepts are -5/4

Solution

y = -3x²+3x-2

It opens downwards as the cofficient of x^2 is negative

Now the vertex for equation ax² + bx + c is given by (-b/2a,f(-b/2a))

And so a = -3, b = 3, c = -2 in our case

We get vertex = (-(-3)/6, f(3/6)) = (1/2, f(1/2))

Now f(1/2) = -3(1/2)²+3(1/2)-2 = -3(1/4)+3/2-2 = -11/4+3/2 = -5/4

hence vertex is (1/2,-5/4) and Also as this is parabola opening downwards so this will be maximum point of parabola

Domain : Since this is polynomial function and so domain is all Real numbers

Range: All values less than or equal to -5/4 and hence (-infinity, -5/4]

Axis of symmetry: x = -b/2a = 1/2

Hence x=1/2 is axis of symmetry

x intercepts: None

y intercepts: (0, -2)

Note: You were incorrect for axis if symmetry, domain , range and y intercept