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

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

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vertices y=x^2+3x-4

vertices

y = 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})

f (x) = - x^{2} + 6 x - 5

f (x) = x^{2} - 5 x + 6

- x^{2} + 4 x - 3

y = 2 x^{2} - 1 - (5 x - 6)

y = x^{2} - x - 20

What is the vertices y=x^2+3x-4 ?
The vertices y=x^2+3x-4 is Minimum (-3/2 ,-25/4)