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

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 = (x - 8)^{2}

y = x^{2} - 8 x + 12

f (x) = (x + 2) (x + 3)

y^{2} = 6 x - 4 y - 12

y = + x^{2} + 8

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