What is the vertices 4x^2-24x+31 ?

The vertices 4x^2-24x+31 is Minimum (3,-5)

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vertices 4x^2-24x+31

vertices

4 x^{2} - 24 x + 31

Minimum (3, - 5)

Solution steps

y = 4 x^{2} - 24 x + 31

The parabola parameters are:

a = 4, b = - 24, c = 31

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

x_{v} = - \frac{( - 24 )}{2 \cdot 4}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(3, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (3, - 5)

y = x^{2} + x

y = 9 x^{2}

y = 8 x^{2} + 22 x - 13

y = x^{2} + 8 x + 18

x^{2} - 5 x - 6