What is the vertices y=7x^2+5 ?

The vertices y=7x^2+5 is Minimum (0,5)

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vertices y=7x^2+5

vertices

y = 7 x^{2} + 5

Minimum (0, 5)

Solution steps

y = 7 x^{2} + 5

The parabola parameters are:

a = 7, b = 0, c = 5

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

x_{v} = - \frac{0}{2 \cdot 7}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(0, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 7

Minimum (0, 5)

- 2 x^{2} - 4 x + 3

y = x^{2} - 6 x - 3

y = - x^{2} - 6 x + 1

y = 3 (x - 1)^{2} - 2

y = - x^{2} + 2 x - 5