What is the vertices 4x^2-16x+11 ?

The vertices 4x^2-16x+11 is Minimum (2,-5)

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vertices 4x^2-16x+11

vertices

4 x^{2} - 16 x + 11

Minimum (2, - 5)

Solution steps

y = 4 x^{2} - 16 x + 11

The parabola parameters are:

a = 4, b = - 16, c = 11

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} = - 5

Therefore the parabola vertex is

(2, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (2, - 5)

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

5 x^{2} + 7 y = 3

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

f (x) = 2 x^{2} - 80 x + 900

y = 5 x^{2}