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

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

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vertices y=2x^2-7

vertices

y = 2 x^{2} - 7

Minimum (0, - 7)

Solution steps

y = 2 x^{2} - 7

The parabola parameters are:

a = 2, b = 0, c = - 7

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

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

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 = 2

Minimum (0, - 7)

y = - 3 x^{2} - 36 x - 104

f (x) = x^{2} - 6 x + 11

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

f (x) = x^{2} + 10 x + 21

y = (9 x - 2) x 3