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

The vertices y=x^2+3x is Minimum (-3/2 ,-9/4)

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vertices y=x^2+3x

vertices

y = x^{2} + 3 x

Minimum (- \frac{3}{2}, - \frac{9}{4})

Solution steps

y = x^{2} + 3 x

The parabola parameters are:

a = 1, b = 3, c = 0

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{9}{4}

Therefore the parabola vertex is

(- \frac{3}{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{3}{2}, - \frac{9}{4})

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

x = - 3 y^{2} + 5

f (x) = x^{2} + 10 x + 6

- 3 x^{2} - 12 x - 9

y = x^{2} - 2 x 5