What is the vertices x^2+3x+1 ?

The vertices x^2+3x+1 is Minimum (-3/2 ,-5/4)

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vertices x^2+3x+1

vertices

x^{2} + 3 x + 1

Minimum (- \frac{3}{2}, - \frac{5}{4})

Solution steps

y = x^{2} + 3 x + 1

The parabola parameters are:

a = 1, b = 3, c = 1

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

x_{v} = - \frac{3}{2 \cdot 1}

Simplify

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = - \frac{5}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{5}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, - \frac{5}{4})

x^{2} + 3 x - 1

x^{2} - 7 x - 8

0 = (y - 1)^{2} - 16 (x + 4)

y = + 8 - 9 x^{2}

2 π r^{2}