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

The vertices f(x)=x^2+x-2 is Minimum (-1/2 ,-9/4)

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

vertices

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

Minimum (- \frac{1}{2}, - \frac{9}{4})

Solution steps

y = x^{2} + x - 2

The parabola parameters are:

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

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

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

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{9}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{1}{2}, - \frac{9}{4})

- 2 x^{2} + 12 x - 14

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

- 2 x^{2} + 12 x

f (x) = 4 x^{2} - 7 x - 2

2 y^{2} + 8 y + x + 1 = 0

What is the vertices f(x)=x^2+x-2 ?
The vertices f(x)=x^2+x-2 is Minimum (-1/2 ,-9/4)