What is the vertices y=2x^2+15x ?

The vertices y=2x^2+15x is Minimum (-15/4 ,-225/8)

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vertices y=2x^2+15x

vertices

y = 2 x^{2} + 15 x

Minimum (- \frac{15}{4}, - \frac{225}{8})

Solution steps

y = 2 x^{2} + 15 x

The parabola parameters are:

a = 2, b = 15, c = 0

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

x_{v} = - \frac{15}{2 \cdot 2}

Simplify

x_{v} = - \frac{15}{4}

Plug in

x_{v} = - \frac{15}{4}

to find the

y_{v}

value

y_{v} = - \frac{225}{8}

Therefore the parabola vertex is

(- \frac{15}{4}, - \frac{225}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{15}{4}, - \frac{225}{8})

f (x) = - x^{2} + 10 x - 16

f (x) = - x^{2} + 5 x - 6

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

y = 3 x^{2} - 3 x - 6

f (x) = x^{2} - 3 x + 10

What is the vertices y=2x^2+15x ?
The vertices y=2x^2+15x is Minimum (-15/4 ,-225/8)