What is the vertices y=x^2+3x+4 ?

The vertices y=x^2+3x+4 is Minimum (-3/2 , 7/4)

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vertices y=x^2+3x+4

vertices

y = x^{2} + 3 x + 4

Minimum (- \frac{3}{2}, \frac{7}{4})

Solution steps

y = x^{2} + 3 x + 4

The parabola parameters are:

a = 1, b = 3, c = 4

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

x_{v} = - \frac{3}{2 \cdot 1}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{7}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, \frac{7}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, \frac{7}{4})

y = 24 x - 3 x^{2}

(x - 5)^{2} = 4 (y - 3)

(x - 4)^{2} = 8 y + 16

- 4 x^{2}

x^{2} - 4 x - 7

What is the vertices y=x^2+3x+4 ?
The vertices y=x^2+3x+4 is Minimum (-3/2 , 7/4)