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

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

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

vertices

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

Minimum (1, - 4)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 3

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 4)

y + x^{2} = 4 x

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

x^{2} = 8 y

y^{2} = 32 x

16 x^{2} - 9 y^{2} = 144

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