1. ## Completing the square

I do not understand this at all. Click image to enlarge . I keep getting the answer $5(t-8/5)^2+1/5$, which means that the vertex of this equation is NOT $(8/5,-64/25)$. I even used h: $8/5$ to plug it into the entire equation $5t^2-16t+13$.

2. ## Re: Completing the square

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

-Dan

3. ## Re: Completing the square

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

4. ## Re: Completing the square

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

5. ## Re: Completing the square

Originally Posted by Eraser147
I do not understand this at all. Click image to enlarge . I keep getting the answer $5(t-8/5)^2+1/5$, which means that the vertex of this equation is NOT $(8/5,-64/25)$. I even used h: $8/5$ to plug it into the entire equation $5t^2-16t+13$.
\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 \begin{align*} \left( \frac{8}{5}, -\frac{64}{5} \right) \end{align*}.

6. ## 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.

7. ## Re: Completing the square

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.

8. ## 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.