What is the vertices-x^2+5x-7 ?

The vertices-x^2+5x-7 is Maximum (5/2 ,-3/4)

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vertices-x^2+5x-7

vertices

- x^{2} + 5 x - 7

Maximum (\frac{5}{2}, - \frac{3}{4})

Solution steps

y = - x^{2} + 5 x - 7

The parabola parameters are:

a = - 1, b = 5, c = - 7

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

x_{v} = - \frac{5}{2 ( - 1 )}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{3}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{3}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{5}{2}, - \frac{3}{4})

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

y = - \frac{1}{16} (x + 2)^{2} - 6

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

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

y = 4 x^{2} - 12 x + 10