What is the vertices y=3-x^2 ?

The vertices y=3-x^2 is Maximum (0,3)

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vertices y=3-x^2

vertices

y = 3 - x^{2}

Maximum (0, 3)

Solution steps

y = 3 - x^{2}

Rewrite

y = 3 - x^{2}

in the form

y = a x^{2} + b x + c

y = - x^{2} + 3

The parabola parameters are:

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

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

x_{v} = - \frac{0}{2 ( - 1 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(0, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 3)

y = 25 x^{2} - 36

f (x) = x^{2} - 4 x + 4

y^{2} = 4 x

y = x^{2} - 2 x - 3

y = x^{2} + 2 x + 1