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What is the positive value of $k$ for which the graph $y=x^2-2kx +16$ is tangent to the x-axis?

My approach to solving this question is to use vertex form to find the value of $x$. Once I have that, I can substitute that value to find the value of $k$, which is $4$.

If I try to apply that same method to this question, enter image description here

I get $p=4\sqrt{3}$. The correct answer is $p = 6$. What is wrong with the way I applied my method?

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In the second problem, the graph is not tangent to the $x$-axis. Instead, since the lowest point of the rope is at a height of $1$, the graph is tangent to the line $y=1$. You could use the same method to solve this problem, but you would need to subtract $1$ from the quadratic first to make it tangent to the $x$-axis.

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  • $\begingroup$ Thank you so much, I can't believe i miss that vital information. I was beginning to question my approach was ever right in the first place @ Eric Wofsey $\endgroup$ – aimepie Nov 23 '17 at 18:36
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Since you haven't provided your solution, I presume you computed the value of $k$ by completing squares. In particular, you transformed $$y = x^2-2kx+16=(x-k)^2+(16-k^2),$$ and put $y=0$ at $x=k$ to find $k=4$.

For the given problem, $$y = \frac{x}{3}(x-p)+4 = \frac{\left(x-\frac{p}{2}\right)^2}{3}+4-\frac{p^2}{12}.$$ Now, for $x=\frac{p}{2}$, $y=1 \implies 4-\frac{p^2}{12}=1 \implies p^2=36 \implies p=6$.

You could also identify $p$ by noting that $y$ achieves its minimum when $x=3$. That is, $\frac{p}{2}=3$.

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  • $\begingroup$ I used the vertex formula -b/2a to find x and back into the original equation. The first part, the lowest point is tangent the x-axis which makes y = 0, for the second part, y is equal to 1. All and all the method works for both problem but I missed the information about y =1 for the second part. Thankyou for taking the time to help though @ Math Lover $\endgroup$ – aimepie Nov 23 '17 at 18:50

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