What is the vertices x^2+5x+1 ?

The vertices x^2+5x+1 is Minimum (-5/2 ,-21/4)

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vertices x^2+5x+1

vertices

x^{2} + 5 x + 1

Minimum (- \frac{5}{2}, - \frac{21}{4})

Solution steps

y = x^{2} + 5 x + 1

The parabola parameters are:

a = 1, b = 5, c = 1

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{21}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{21}{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{21}{4})

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

f (x) = 2 x^{2} - 4 x + 1

y = - x^{2} + 10 x

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

+ 1 + x^{2}