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

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

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

vertices

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

Minimum (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 \cdot 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

Minimum (2, - 1)

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

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

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

y^{2} = - 25 x

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