What is the vertices y=x^2+7 ?

The vertices y=x^2+7 is Minimum (0,7)

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vertices y=x^2+7

vertices

y = x^{2} + 7

Minimum (0, 7)

Solution steps

y = x^{2} + 7

The parabola parameters are:

a = 1, b = 0, c = 7

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} = 7

Therefore the parabola vertex is

(0, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 7)

f (x) = \frac{3}{x} - 3

f (x) = 2 x + 9 - 4

f (x) = \frac{3 x ^{2} - 5 x - 2}{x + 3}

f (x) = x^{2} - 1, x \leq 0

f (x) = - \frac{2}{3} csc (2 x)