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

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

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

vertices

y = 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})

x^{2} + 5 x + 6

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

y^{2} = 12 x

x^{2} + 6 x + 5

y = 3 x^{2} - 5 x 7

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