What is the vertices 8x^2+12x+3 ?

The vertices 8x^2+12x+3 is Minimum (-3/4 ,-3/2)

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vertices 8x^2+12x+3

vertices

8 x^{2} + 12 x + 3

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

Solution steps

y = 8 x^{2} + 12 x + 3

The parabola parameters are:

a = 8, b = 12, c = 3

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

x_{v} = - \frac{12}{2 \cdot 8}

Simplify

- \frac{12}{2 \cdot 8} : - \frac{3}{4}

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{3}{2}

Therefore the parabola vertex is

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

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 8

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

- 5 x^{2} + 118 x + 7

y^{2} = - 6 x

4 x - 2 x^{2}

y = \frac{95 x ^{2} - 10 x + 26}{10}

(x - 1)^{2} = - 2 y - 11

What is the vertices 8x^2+12x+3 ?
The vertices 8x^2+12x+3 is Minimum (-3/4 ,-3/2)