What is the vertices x^2-3x-4 ?

The vertices x^2-3x-4 is Minimum (3/2 ,-25/4)

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vertices x^2-3x-4

vertices

x^{2} - 3 x - 4

Minimum (\frac{3}{2}, - \frac{25}{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{25}{4}

Therefore the parabola vertex is

(\frac{3}{2}, - \frac{25}{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{25}{4})

4 x^{2} - 8 x - 3 y - 2 = 0

y = - \frac{( x + 3 ) ^{2}}{20} + 1

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

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

y = x^{2} - 5 x 3