What is the vertices f(x)=x^2-3x+10 ?

The vertices f(x)=x^2-3x+10 is Minimum (3/2 , 31/4)

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vertices f(x)=x^2-3x+10

vertices

f (x) = x^{2} - 3 x + 10

Minimum (\frac{3}{2}, \frac{31}{4})

Solution steps

y = x^{2} - 3 x + 10

The parabola parameters are:

a = 1, b = - 3, c = 10

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{31}{4}

Therefore the parabola vertex is

(\frac{3}{2}, \frac{31}{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{31}{4})

- 2 x^{2} + 12 x - 20

y^{2} + 2 x + 6 y + 17 = 0

f (x) = 2 x^{2} + 3 x + 1

- (x + 2)^{2} + 6

f (x) = 2 x^{2} + 3 x - 5

What is the vertices f(x)=x^2-3x+10 ?
The vertices f(x)=x^2-3x+10 is Minimum (3/2 , 31/4)