What is the vertices f(x)=4-x^2 ?

The vertices f(x)=4-x^2 is Maximum (0,4)

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vertices f(x)=4-x^2

vertices

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

Maximum (0, 4)

Solution steps

y = 4 - x^{2}

Rewrite

y = 4 - x^{2}

in the form

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

y = - x^{2} + 4

The parabola parameters are:

a = - 1, b = 0, c = 4

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} = 4

Therefore the parabola vertex is

(0, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (0, 4)

f (x) = x^{2} - 4 x + 7

y = x^{2} - 6 x + 10

y = 3 x^{2} - 30 x + 77

y = x^{2} - 4

y = x^{2} + 1