What is the vertices x^2+2x-1-(2x^2-3x+6) ?

The vertices x^2+2x-1-(2x^2-3x+6) is Maximum (5/2 ,-3/4)

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vertices x^2+2x-1-(2x^2-3x+6)

vertices

x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

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

Solution steps

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

Rewrite

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

in the form

y = a x^{2} + b x + c

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})

3 x^{2} - x - 2

f (x) = 4 x^{2} + 2

x^{2} + 3

(2 x - 1) (5 x - 6)

x^{2} + 2 x + 1

What is the vertices x^2+2x-1-(2x^2-3x+6) ?
The vertices x^2+2x-1-(2x^2-3x+6) is Maximum (5/2 ,-3/4)