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})

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

y^{2} - 4 y + 4 x = - 16

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

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

\frac{y ^{2}}{25} - \frac{x ^{2}}{16} = 1

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