What is the vertices f(x)=x^2+5x+6 ?

The vertices f(x)=x^2+5x+6 is Minimum (-5/2 ,-1/4)

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vertices f(x)=x^2+5x+6

vertices

f (x) = x^{2} + 5 x + 6

Minimum (- \frac{5}{2}, - \frac{1}{4})

Solution steps

y = x^{2} + 5 x + 6

The parabola parameters are:

a = 1, b = 5, c = 6

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

x_{v} = - \frac{5}{2 \cdot 1}

Simplify

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{5}{2}, - \frac{1}{4})

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

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

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

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63

y^{2} = 28 x

What is the vertices f(x)=x^2+5x+6 ?
The vertices f(x)=x^2+5x+6 is Minimum (-5/2 ,-1/4)