What is the vertices f(x)=4x^2-8x+7 ?

The vertices f(x)=4x^2-8x+7 is Minimum (1,3)

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vertices f(x)=4x^2-8x+7

vertices

f (x) = 4 x^{2} - 8 x + 7

Minimum (1, 3)

Solution steps

y = 4 x^{2} - 8 x + 7

The parabola parameters are:

a = 4, b = - 8, c = 7

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

x_{v} = - \frac{( - 8 )}{2 \cdot 4}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (1, 3)

f (x) = 2 x^{2} + 5 x + 2

f (x) = 8 x^{2}

y = 3 x - x^{2}

y = 3 x^{2} + 12 x - 3

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

What is the vertices f(x)=4x^2-8x+7 ?
The vertices f(x)=4x^2-8x+7 is Minimum (1,3)