What is the vertices y=x^2+4x+7 ?

The vertices y=x^2+4x+7 is Minimum (-2,3)

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vertices y=x^2+4x+7

vertices

y = x^{2} + 4 x + 7

Minimum (- 2, 3)

Solution steps

y = x^{2} + 4 x + 7

The parabola parameters are:

a = 1, b = 4, c = 7

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, 3)

- x^{2} - 4 x + 5

f (x) = x^{2} + 16 x + 64

y = 2 x^{2} - 4 x + 5

y = 6 x^{2} - 1

(x + 5)^{2} = - 20 (y - 2)