What is the vertices f(x)=-x^2-4x-3 ?

The vertices f(x)=-x^2-4x-3 is Maximum (-2,1)

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vertices f(x)=-x^2-4x-3

vertices

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

Maximum (- 2, 1)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 4, c = - 3

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} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 2, 1)

x^{2} + y^{2} - 16 = 0

\frac{y ^{2}}{25} - \frac{x ^{2}}{4} = 1

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

x^{2} = - 16 y

x^{2} - 6 x - 4 y + 5 = 0

What is the vertices f(x)=-x^2-4x-3 ?
The vertices f(x)=-x^2-4x-3 is Maximum (-2,1)