Solving for x in P(x) = √6+5x-x2

[Wa1337]

Homework Statement

P(x) = √6+5x-x²

Homework Equations

The Attempt at a Solution

P(x) = √6+5x-x²

-x²+5x+6 >/= 0
x²-5x-6 </= 0
(x-6)(x+1) </= 0

 

[SammyS]

Do you mean $P(x) =\sqrt{6+5x-x^2}\,?$

If so, then use parentheses to make things clear.

P(x) = √(6+5x-x²).

Otherwise it looks like you have $P(x) =(\sqrt{6})+5x-x^2\,?$

Homework Statement

P(x) = √6+5x-x²

Homework Equations

The Attempt at a Solution

P(x) = √6+5x-x²

-x²+5x+6 >/= 0
x²-5x-6 </= 0
(x-6)(x+1) </= 0

What you have done will allow you to find the domain of P(x).

To find the range, find the vertex of the parabola of f(x) = 6+5x-x² . Then proceed from there.

 

[Wa1337]

In vertex form it is y=(x-2.5)²-12.25