What is the vertices f(x)=x^2+5x+4 ?

The vertices f(x)=x^2+5x+4 is Minimum (-5/2 ,-9/4)

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vertices f(x)=x^2+5x+4

vertices

f (x) = x^{2} + 5 x + 4

Minimum (- \frac{5}{2}, - \frac{9}{4})

Solution steps

y = x^{2} + 5 x + 4

The parabola parameters are:

a = 1, b = 5, c = 4

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{9}{4}

Therefore the parabola vertex is

(- \frac{5}{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{5}{2}, - \frac{9}{4})

+ y^{2} + 12 x + 4 y - 32 = 0

f (x) = - 2 x^{2} - 16 x - 30

x = - 5 y^{2}

y = 6 x^{2} - 15 x

f (x) = - (x - 5)^{2} - 4

What is the vertices f(x)=x^2+5x+4 ?
The vertices f(x)=x^2+5x+4 is Minimum (-5/2 ,-9/4)