What is the vertices y=x^2+4x-3 ?

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

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vertices y=x^2+4x-3

vertices

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

Minimum (- 2, - 7)

Solution steps

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

The parabola parameters are:

a = 1, b = 4, c = - 3

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} = - 7

Therefore the parabola vertex is

(- 2, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 7)

f (x) = x^{2} - 12 x + 27

3 x^{2} + 2

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

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

- x^{2} - 4 x + 5