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

The vertices f(x)=x^2-3x-10 is Minimum (3/2 ,-49/4)

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

vertices

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

Minimum (\frac{3}{2}, - \frac{49}{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{49}{4}

Therefore the parabola vertex is

(\frac{3}{2}, - \frac{49}{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{49}{4})

- x^{2} + 2 x - 10

f (x) = 3 x^{2} - 12 x + 13

y = x^{2} - x - 6

f (x) = - x^{2} + 6 x - 8

- x^{2} + 5 x - 7

What is the vertices f(x)=x^2-3x-10 ?
The vertices f(x)=x^2-3x-10 is Minimum (3/2 ,-49/4)