What is the vertices f(x)=-(x^2+3x-4) ?

The vertices f(x)=-(x^2+3x-4) is Maximum (-3/2 , 25/4)

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vertices f(x)=-(x^2+3x-4)

vertices

f (x) = - (x^{2} + 3 x - 4)

Maximum (- \frac{3}{2}, \frac{25}{4})

Solution steps

y = - (x^{2} + 3 x - 4)

Rewrite

y = - (x^{2} + 3 x - 4)

in the form

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

y = - x^{2} - 3 x + 4

The parabola parameters are:

a = - 1, b = - 3, c = 4

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{25}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, \frac{25}{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{25}{4})

y = 4 x^{2} - 16

f (x) = x^{2} - 8 x + 17

y = 24000 x^{2} + 900 x - 150

16 - (12 x + 8)^{2}

y^{2} - 8 x - 6 y - 23 = 0

What is the vertices f(x)=-(x^2+3x-4) ?
The vertices f(x)=-(x^2+3x-4) is Maximum (-3/2 , 25/4)