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

The vertices y=(x-3)^2-2 is Minimum (3,-2)

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

vertices

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

Minimum (3, - 2)

Solution steps

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

Rewrite

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

in the form

y = a x^{2} + b x + c

y = x^{2} - 6 x + 7

The parabola parameters are:

a = 1, b = - 6, c = 7

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 2)

x^{2} - 5 x - 36

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

100 (1 - \frac{x}{40})^{2}

y = x^{2} + 12 x + 16

f (x) = - 5 x^{2} + 20 x + 60