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

The vertices x^2+3x-1 is Minimum (-3/2 ,-13/4)

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vertices x^2+3x-1

vertices

Minimum (- \frac{3}{2}, - \frac{13}{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{13}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{13}{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{13}{4})