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

The vertices f(x)=-x^2+2x+3 is Maximum (1,4)

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

vertices

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

Maximum (1, 4)

Solution steps

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

The parabola parameters are:

a = - 1, b = 2, c = 3

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

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

Simplify

- \frac{2}{2 ( - 1 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 4)

f (x) = 4 x^{2}

2 x^{2} + 2 x + 3 y = 0

y = 4 x - x^{2}

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

x^{2} + 5 x + 6

What is the vertices f(x)=-x^2+2x+3 ?
The vertices f(x)=-x^2+2x+3 is Maximum (1,4)