What is the vertices f(x)=-3x^2+1 ?

The vertices f(x)=-3x^2+1 is Maximum (0,1)

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vertices f(x)=-3x^2+1

vertices

f (x) = - 3 x^{2} + 1

Maximum (0, 1)

Solution steps

y = - 3 x^{2} + 1

The parabola parameters are:

a = - 3, b = 0, c = 1

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{0}{2 ( - 3 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(0, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (0, 1)

f (x) = (x - 1)^{2} + 2

f (x) = x^{2} + 12 x + 16

x^{2} + 4 x - 5

- x^{2} + 3 x - 5

y = - x^{2} + 8 x