What is the vertices 10+3x^2 ?

The vertices 10+3x^2 is Minimum (0,10)

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vertices 10+3x^2

vertices

10 + 3 x^{2}

Minimum (0, 10)

Solution steps

y = 10 + 3 x^{2}

Rewrite

y = 10 + 3 x^{2}

in the form

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

y = 3 x^{2} + 10

The parabola parameters are:

a = 3, b = 0, c = 10

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

x_{v} = - \frac{0}{2 \cdot 3}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 10

Therefore the parabola vertex is

(0, 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (0, 10)

x^{2} - 5 x - 24

y = (2 x)^{2}

y = 2 x^{2} - 19 x + 9

4 x^{2} - 20 x - 24 y + 97 = 0

y - 2 = (x - 1)^{2}