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

The vertices f(x)=x^2+2x-3 is Minimum (-1,-4)

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

vertices

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

Minimum (- 1, - 4)

Solution steps

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

The parabola parameters are:

a = 1, b = 2, c = - 3

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 4)

- (x + 3)^{2} - 1

49 x^{2} + 16 y^{2} = 784

3 x^{2}

y^{2} - 8 x + 2 y + 17 = 0

x^{2} = - 36 y

What is the vertices f(x)=x^2+2x-3 ?
The vertices f(x)=x^2+2x-3 is Minimum (-1,-4)