What is the vertices f(x)=x^2+4x-3 ?

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

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vertices f(x)=x^2+4x-3

vertices

f (x) = 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} + 10 x + 9

y = - 3 x^{2} - 12 x - 9

x^{2} = 8 (y + 4)

y = 3 x^{2} - 12 x + 8

f (x) = - x^{2} - 4 x + 1

What is the vertices f(x)=x^2+4x-3 ?
The vertices f(x)=x^2+4x-3 is Minimum (-2,-7)