What is the vertices f(x)=-x^2+3x-5 ?

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

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vertices f(x)=-x^2+3x-5

vertices

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

Maximum (\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 ( - 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

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

y = x^{2} - 10 x - 4

x^{2} + 6 y = 0

f (x) = 7 x^{2} - x + 1

f (x) = 3 x^{2} - 6 x + 2

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

What is the vertices f(x)=-x^2+3x-5 ?
The vertices f(x)=-x^2+3x-5 is Maximum (3/2 ,-11/4)