What is the vertices f(x)=x^2+x-6 ?

The vertices f(x)=x^2+x-6 is Minimum (-1/2 ,-25/4)

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vertices f(x)=x^2+x-6

vertices

f (x) = x^{2} + x - 6

Minimum (- \frac{1}{2}, - \frac{25}{4})

Solution steps

y = x^{2} + x - 6

The parabola parameters are:

a = 1, b = 1, c = - 6

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{25}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{25}{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{25}{4})

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

x^{2} + 9 y^{2} = 9

f (x) = x^{2} + 4 x - 12

x^{2} - 10 x + y^{2} + 6 y - 1 = 0

What is the vertices f(x)=x^2+x-6 ?
The vertices f(x)=x^2+x-6 is Minimum (-1/2 ,-25/4)