What is the vertices y=2x(x-4)+7 ?

The vertices y=2x(x-4)+7 is Minimum (2,-1)

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vertices y=2x(x-4)+7

vertices

y = 2 x (x - 4) + 7

Minimum (2, - 1)

Solution steps

y = 2 x (x - 4) + 7

Rewrite

y = 2 x (x - 4) + 7

in the form

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

y = 2 x^{2} - 8 x + 7

The parabola parameters are:

a = 2, b = - 8, c = 7

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

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

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 = 2

Minimum (2, - 1)

x = y^{2} + 2 y - 1 - (2 y^{2} - 3 y + 6)

f (x) = x^{2} + 6 x - 8

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

f (x) = x^{2} + 6 x + 4

f (x) = 2 x^{2} - 10 x + 12