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

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

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vertices f(x)=x^2+4

vertices

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

Minimum (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 \cdot 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

Minimum (0, 4)

\frac{y ^{2}}{25} - \frac{x ^{2}}{16} = 1

\frac{x ^{2}}{169} + \frac{y ^{2}}{25} = 1

2 x - y^{2} = 4

f (x) = x^{2} - 24 x - 12

x^{2} = - 12 y