# Thread: Locate relative extrema is this working correct?

1. ## Locate relative extrema is this working correct?

Locate and describe the relative extrema then sketch the graph of

is the theory correct?

$\displaystyle 3x^2 + 6x - 9 = 0$

$\displaystyle 3(x^2 + 2x - 3) = 0$

$\displaystyle 3(x + 3)(x - 1) = 0$

3. After $\displaystyle 3x^2 + 6x - 9 = 0$, you wrote $\displaystyle 3x(x - 1) = 0$, which is false. Expand the left-hand side of the false equation to see why.

Instead, divide the equation $\displaystyle 3x^2 + 6x - 9 = 0$ by $\displaystyle 3$:

$\displaystyle x^2 + 2x - 3= 0$

Then, factor the right-hand side of the equation:

$\displaystyle (x - 1)(x + 3) = 0$

Set each factor on the right-hand side of the equation equal to 0.

$\displaystyle x - 1 = 0$ or $\displaystyle x + 3 = 0$

Solve each equation in terms of x. The solutions are the critical values.

4. Originally Posted by NOX Andrew
After $\displaystyle 3x^2 + 6x - 9 = 0$, you wrote $\displaystyle 3x(x - 1) = 0$, which is false. Expand the left-hand side of the false equation to see why.

Instead, divide the equation $\displaystyle 3x^2 + 6x - 9 = 0$ by $\displaystyle 3$:

$\displaystyle x^2 + 2x - 3= 0$

Then, factor the right-hand side of the equation:

$\displaystyle (x - 1)(x + 3) = 0$

Set each factor on the right-hand side of the equation equal to 0.

$\displaystyle x - 1 = 0$ or $\displaystyle x + 3 = 0$

Solve each equation in terms of x. The solutions are the critical values.

so x=1 or x =-3

what do i do with the 3 outside the brackets? 3x = 0 ?

5. No, only set factors with an $\displaystyle x$ equal to 0. In other words, don't multiply a constant by an $\displaystyle x$ and then set the product equal to 0. If you really wanted to, you could set just $\displaystyle 3$ equal to $\displaystyle 0$. However, $\displaystyle 3 = 0$ is false, so there are no solutions to that specific equation.