What is the vertices y=4x^2-16x+22 ?

The vertices y=4x^2-16x+22 is Minimum (2,6)

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vertices y=4x^2-16x+22

vertices

y = 4 x^{2} - 16 x + 22

Minimum (2, 6)

Solution steps

y = 4 x^{2} - 16 x + 22

The parabola parameters are:

a = 4, b = - 16, c = 22

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

x_{v} = - \frac{( - 16 )}{2 \cdot 4}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(2, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (2, 6)

x^{2} + 8 x - 12 y + 88 = 0

4 x^{2} - 16 x + 11

f (x) = x^{2} - 5 x - 6

5 x^{2} + 7 y = 3

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

What is the vertices y=4x^2-16x+22 ?
The vertices y=4x^2-16x+22 is Minimum (2,6)