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

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

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

vertices

f (x) = - 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)

3 y + 4 x = - 2 x^{2} - 14

x^{2} = 20 y

\frac{x ^{2}}{169} + \frac{y ^{2}}{25} = 1

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

2 x^{2} - 6 x + 2 y^{2} + 2 y = 45

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