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

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

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vertices y=x^2+2x-3

vertices

y = 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)

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

2 x^{2}

4 y^{2} - 25 x^{2} = 100

- 9 x^{2} + y^{2} - 72 x - 153 = 0

f (x) = x^{2} - 2 x - 15