What is the vertices y=x^2+3x-2 ?

The vertices y=x^2+3x-2 is Minimum (-3/2 ,-17/4)

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vertices y=x^2+3x-2

vertices

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

Minimum (- \frac{3}{2}, - \frac{17}{4})

Solution steps

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

The parabola parameters are:

a = 1, b = 3, c = - 2

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{17}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{17}{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{17}{4})

100 (1 - \frac{x}{40})^{2}

y = x^{2} + 12 x + 16

f (x) = - 5 x^{2} + 20 x + 60

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

f (x) = - x^{2} - 14 x - 42

What is the vertices y=x^2+3x-2 ?
The vertices y=x^2+3x-2 is Minimum (-3/2 ,-17/4)