What is the vertices 4(x+6)+3(2+x+3x^2)-2x^2 ?

The vertices 4(x+6)+3(2+x+3x^2)-2x^2 is Minimum (-1/2 , 113/4)

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vertices 4(x+6)+3(2+x+3x^2)-2x^2

vertices

4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

Minimum (- \frac{1}{2}, \frac{113}{4})

Solution steps

y = 4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

Rewrite

y = 4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

in the form

y = a x^{2} + b x + c

y = 7 x^{2} + 7 x + 30

The parabola parameters are:

a = 7, b = 7, c = 30

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

x_{v} = - \frac{7}{2 \cdot 7}

Simplify

- \frac{7}{2 \cdot 7} : - \frac{1}{2}

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = \frac{113}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{113}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 7

Minimum (- \frac{1}{2}, \frac{113}{4})

f (x) = - x^{2} + 6 x - 11

y^{2} = - 20 x

2 x^{2} - 6 x + 15

(x + 1)^{2} = - 6 (y - 4)

x^{2} - 2 x - 4 y + 13 = 0

What is the vertices 4(x+6)+3(2+x+3x^2)-2x^2 ?
The vertices 4(x+6)+3(2+x+3x^2)-2x^2 is Minimum (-1/2 , 113/4)