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

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

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vertices-x^2+4x-3

vertices

- x^{2} + 4 x - 3

Maximum (2, 1)

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} = 1

Therefore the parabola vertex is

(2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 1)

y = 2 x^{2} - 1 - (5 x - 6)

y = x^{2} - x - 20

\frac{- 3 x ^{2} - 2 x + 6}{5}

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

f (x) = 4 x^{2}