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

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

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vertices+7x^2+5

vertices

+ 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)

y = 6 x^{2} - 3 x 5

3 x^{2} + 5 x - 12

(x - 1) (x - 3)

y = x^{2} - 8 x + 15

y^{2} = - 28 x