What is the vertices f(x)=2x^2+6x+1 ?

The vertices f(x)=2x^2+6x+1 is Minimum (-3/2 ,-7/2)

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vertices f(x)=2x^2+6x+1

vertices

f (x) = 2 x^{2} + 6 x + 1

Minimum (- \frac{3}{2}, - \frac{7}{2})

Solution steps

y = 2 x^{2} + 6 x + 1

The parabola parameters are:

a = 2, b = 6, c = 1

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

x_{v} = - \frac{6}{2 \cdot 2}

Simplify

- \frac{6}{2 \cdot 2} : - \frac{3}{2}

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{7}{2}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{7}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{3}{2}, - \frac{7}{2})

x^{2} + 10 x + 11

y = x^{2} + 2 x - 35

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

f (x) = x^{2} - 10 x + 9

y = - x^{2} - 8

What is the vertices f(x)=2x^2+6x+1 ?
The vertices f(x)=2x^2+6x+1 is Minimum (-3/2 ,-7/2)