Re: Completing the square

Your completing the square work is correct. (The answer in the "Click to enlarge" panel, that is.)

-Dan

Re: Completing the square

Why? Why is the k value not 1/5?

Re: Completing the square

Can you show me the steps for the solution so I can look at the mechanics of it?

Re: Completing the square

Quote:

Originally Posted by

**Eraser147** I do not understand this at all. Click image to enlarge

Attachment 25278. I keep getting the answer $\displaystyle 5(t-8/5)^2+1/5$, which means that the vertex of this equation is NOT $\displaystyle (8/5,-64/25)$. I even used h:$\displaystyle 8/5$ to plug it into the entire equation $\displaystyle 5t^2-16t+13$.

$\displaystyle \displaystyle \begin{align*} 5 \left[ \left( t - \frac{8}{5} \right) ^2 - \frac{64}{25} \right] &= 5 \cdot \left( t - \frac{8}{5} \right)^2 + 5 \cdot \left( -\frac{64}{25} \right) \\ &= 5\left( t - \frac{8}{5} \right)^2 - \frac{64}{5} \end{align*}$

So your turning point should be $\displaystyle \displaystyle \begin{align*} \left( \frac{8}{5}, -\frac{64}{5} \right) \end{align*}$.

Re: Completing the square

It wouldn't make sense because if you plugin 8/5 into the original equation, 1/5 IS the answer for the k value.

Re: Completing the square

Quote:

Originally Posted by

**Eraser147** It wouldn't make sense because if you plugin 8/5 into the original equation, 1/5 IS the answer for the k value.

No amount of arguing is going to make your answer correct. You correctly completed the square, you did NOT do the expansion correctly.

Re: Completing the square

No no no no, just no. That image is what the answer the solution manual has. THAT's NOT my answer. My answer is with the 1/5. I know how to expand, but no matter how I calculate it, it won't show the k value to have 1/5. Again, it is easily testable by jut plugging in k into the original polynomial above.