What is the vertices-x^2+4x+3 ?

The vertices-x^2+4x+3 is Maximum (2,7)

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vertices-x^2+4x+3

vertices

- x^{2} + 4 x + 3

Maximum (2, 7)

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{4}{2 ( - 1 )}

Simplify

- \frac{4}{2 ( - 1 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(2, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 7)

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

(x + 4)^{2}

f (x) = x^{2} - 6 x - 7

x^{2} - 2 x - 8

x^{2} - 6 x + 10