What is the vertices 5+4x-x^2 ?

The vertices 5+4x-x^2 is Maximum (2,9)

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vertices 5+4x-x^2

vertices

5 + 4 x - x^{2}

Maximum (2, 9)

Solution steps

y = 5 + 4 x - x^{2}

Rewrite

y = 5 + 4 x - x^{2}

in the form

y = a x^{2} + b x + c

y = - x^{2} + 4 x + 5

The parabola parameters are:

a = - 1, b = 4, c = 5

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} = 9

Therefore the parabola vertex is

(2, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (2, 9)

8 - 5 x^{2}

f (x) = 2 x^{2} - 4 x - 6

f (x) = 2 x^{2} - 4 x - 3

2 x^{2} - 1 + 5 x

y = - x^{2} + 18 x - 82