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

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

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vertices f(x)=-x^2+4

vertices

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

Maximum (0, 4)

Solution steps

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)

x^{2} + y^{2} = 9

3 x^{2} - 8 y^{2} + 12 x + 16 y + 20 = 0

f (x) = - 3 x^{2} + 6 x - 1

\frac{( y + 2 ) ^{2}}{16} - \frac{( x - 3 ) ^{2}}{9} = 1

f (x) = x^{2} - 2 x - 8