What is the vertices x^2-7x+12 ?

The vertices x^2-7x+12 is Minimum (7/2 ,-1/4)

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vertices x^2-7x+12

vertices

x^{2} - 7 x + 12

Minimum (\frac{7}{2}, - \frac{1}{4})

Solution steps

y = x^{2} - 7 x + 12

The parabola parameters are:

a = 1, b = - 7, c = 12

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

x_{v} = - \frac{( - 7 )}{2 \cdot 1}

Simplify

x_{v} = \frac{7}{2}

Plug in

x_{v} = \frac{7}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(\frac{7}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{7}{2}, - \frac{1}{4})

y = - x^{2} + 1

f (x) = - 2 x^{2} + 20 x + 48

y = x^{2} - 15 x

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

f (x) = (x - 2)^{2} - 1