What is the vertices y=3x^2+30x+65 ?

The vertices y=3x^2+30x+65 is Minimum (-5,-10)

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vertices y=3x^2+30x+65

vertices

y = 3 x^{2} + 30 x + 65

Minimum (- 5, - 10)

Solution steps

y = 3 x^{2} + 30 x + 65

The parabola parameters are:

a = 3, b = 30, c = 65

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

x_{v} = - \frac{30}{2 \cdot 3}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = - 10

Therefore the parabola vertex is

(- 5, - 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 5, - 10)

f (x) = - x^{2} + 3 x + 4

y = (x - 8)^{2}

y = x^{2} - 8 x + 12

f (x) = (x + 2) (x + 3)

y^{2} = 6 x - 4 y - 12

What is the vertices y=3x^2+30x+65 ?
The vertices y=3x^2+30x+65 is Minimum (-5,-10)