What is the vertices f(x)=4x^2-16x+10 ?

The vertices f(x)=4x^2-16x+10 is Minimum (2,-6)

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vertices f(x)=4x^2-16x+10

vertices

f (x) = 4 x^{2} - 16 x + 10

Minimum (2, - 6)

Solution steps

y = 4 x^{2} - 16 x + 10

The parabola parameters are:

a = 4, b = - 16, c = 10

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)

2 x^{2} - 6 x + 4

f (x) = - 3 x^{2} - 18 x + 2

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

y = x^{2} - 4 x + 1

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

What is the vertices f(x)=4x^2-16x+10 ?
The vertices f(x)=4x^2-16x+10 is Minimum (2,-6)