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

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

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

vertices

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

Minimum (4, - 3)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} - 8 x + 13

The parabola parameters are:

a = 1, b = - 8, c = 13

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

x_{v} = - \frac{( - 8 )}{2 \cdot 1}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(4, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 3)

f (x) = 2 x^{2} - 3 x - 5

y = - 2 x^{2} - 16 x - 37

x = 3 y^{2}

y = x^{2} + 4 x - 12

2 (x + 1)^{2} - 3

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