What is the vertices-x^2+3x+1 ?

The vertices-x^2+3x+1 is Maximum (3/2 , 13/4)

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vertices-x^2+3x+1

vertices

- x^{2} + 3 x + 1

Maximum (\frac{3}{2}, \frac{13}{4})

Solution steps

y = - x^{2} + 3 x + 1

The parabola parameters are:

a = - 1, b = 3, c = 1

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

x_{v} = - \frac{3}{2 ( - 1 )}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{13}{4}

Therefore the parabola vertex is

(\frac{3}{2}, \frac{13}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{3}{2}, \frac{13}{4})

f (x) = x^{2} + 12 x + 35

y^{2} + 14 y + 4 x + 45 = 0

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

x^{2} + 8 x + 12

x^{2} + 8 x + 13