What is the vertices f(x)=3x^2-18x+23 ?

The vertices f(x)=3x^2-18x+23 is Minimum (3,-4)

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vertices f(x)=3x^2-18x+23

vertices

f (x) = 3 x^{2} - 18 x + 23

Minimum (3, - 4)

Solution steps

y = 3 x^{2} - 18 x + 23

The parabola parameters are:

a = 3, b = - 18, c = 23

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

x_{v} = - \frac{( - 18 )}{2 \cdot 3}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(3, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (3, - 4)

x^{2} + y^{2} \leq 25

2 x^{2} + y^{2} = 2

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

y = - 4 x^{2}

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

What is the vertices f(x)=3x^2-18x+23 ?
The vertices f(x)=3x^2-18x+23 is Minimum (3,-4)