What is the vertices x^2-4x-7 ?

The vertices x^2-4x-7 is Minimum (2,-11)

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vertices x^2-4x-7

vertices

x^{2} - 4 x - 7

Minimum (2, - 11)

Solution steps

y = x^{2} - 4 x - 7

The parabola parameters are:

a = 1, b = - 4, c = - 7

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 11

Therefore the parabola vertex is

(2, - 11)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 11)

f (x) = - 3 x^{2} + 4 x - 1

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

f (x) = - 9.5 x^{2} - 47.5 x + 63

f (x) = 3 (x - 1) (x + 2)

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x