What is the vertices f(x)=-x^2-4x+1 ?

The vertices f(x)=-x^2-4x+1 is Maximum (-2,5)

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vertices f(x)=-x^2-4x+1

vertices

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

Maximum (- 2, 5)

Solution steps

y = - x^{2} - 4 x + 1

The parabola parameters are:

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

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

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

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(- 2, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 5)

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

x = (y - 3)^{2}

y^{2} = x + 1

f (x) = - x^{2} + 8 x - 12

x^{2} - 6 x + 8

What is the vertices f(x)=-x^2-4x+1 ?
The vertices f(x)=-x^2-4x+1 is Maximum (-2,5)