What is the vertices x^2-3x+5 ?

The vertices x^2-3x+5 is Minimum (3/2 , 11/4)

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vertices x^2-3x+5

vertices

x^{2} - 3 x + 5

Minimum (\frac{3}{2}, \frac{11}{4})

Solution steps

y = x^{2} - 3 x + 5

The parabola parameters are:

a = 1, b = - 3, c = 5

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

x_{v} = - \frac{( - 3 )}{2 \cdot 1}

Simplify

x_{v} = \frac{3}{2}

Plug in

x_{v} = \frac{3}{2}

to find the

y_{v}

value

y_{v} = \frac{11}{4}

Therefore the parabola vertex is

(\frac{3}{2}, \frac{11}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{3}{2}, \frac{11}{4})

3 x^{2} - 3 x

y = 2 x^{2} + 12 x + 17

4 x^{2} + 8 x + 4

y = (x + 2)^{2} + 1

3 x^{2} + 6 x + 2