I'm not really sure how to go about this question yet, I haven't done anything like it before its a first for me any help would be appreciated, thank

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Jeff1801

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If the graph of y=(x-3)(1-x) is tangent to the graph of y=kx^2 then, calculate the value of k.
 
I haven't done that with those two equations, either. How do you think you should proceed? What are the requirements for "is tangent to"? Does it mean they must intersect? Are you SURE there is such a place where they are tangent?
 
tangency between two parabolas implies they “touch” at a single point

[MATH]kx^2 = (x-3)(1-x)[/MATH]
where there is only one solution for x ...
 
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You want the point of intersection to have the same tangent line. Hmm, I would then think that their derivatives would be equal at this point. You need to think!
 
skeeter said:
tangency between two parabolas implies they “touch” at a single point

[MATH]kx^2 = (x-3)(1-x)[/MATH]
where there is only one solution for x ...
Click to expand...
skeeter, can you explain how you know that this can only happen if the two functions have exactly one point of intersection?
I gambled on that being true and it worked. The whole problem took less than one minute. But in my opinion, I was lucky as I hoped (because it made things easier!) that it was exactly when the discriminant was 0. I will also think about it but right now my brain is fried from doing math (so do I do? I hang out on this site)
 
skeeter said:
tangency between two parabolas implies they “touch” at a single point

[MATH]kx^2 = (x-3)(1-x)[/MATH]
where there is only one solution for x ...
Click to expand...
What was your solution for k?

Jomo said:
skeeter, can you explain how you know that this can only happen if the two functions have exactly one point of intersection?
I gambled on that being true and it worked. The whole problem took less than one minute. But in my opinion, I was lucky as I hoped (because it made things easier!) that it was exactly when the discriminant was 0. I will also think about it but right now my brain is fried from doing math (so do I do? I hang out on this site)
Click to expand...
What was your solution for k?
 
Jomo said:
skeeter, can you explain how you know that this can only happen if the two functions have exactly one point of intersection?
Click to expand...
@Jomo
(I shall leave that to skeeter)
However in this case you know it from what you suggested:
Jomo said:
I would then think that their derivatives would be equal at this point.
Click to expand...
You will have found the one value of x for which the two graphs have the same gradient.
When you then solve the quadratic to find the point(s) of intersection, you will discover that the discriminant must be 0, in order to get the above-mentioned value for x.
(k=-1 having to be ruled out separately)
 
Jeff1801 said:
This question is new to me so I still need to learn how to do it. I don't know how to go about getting the value for k.
Click to expand...
Please share your work - so that we can HELP you understand and solve the problem.

If you had followed the discussion above (with pencil|paper ) you surely have some work to share!
 
Jomo said:
skeeter, can you explain how you know that this can only happen if the two functions have exactly one point of intersection?
I gambled on that being true and it worked. The whole problem took less than one minute. But in my opinion, I was lucky as I hoped (because it made things easier!) that it was exactly when the discriminant was 0. I will also think about it but right now my brain is fried from doing math (so do I do? I hang out on this site)
Click to expand...
Some functions/shapes are tangent only when they have a single intersection point. E.g. circles with different radius.
 
Are you studying algebra or calculus. We can’t even begin to help you if you do not tell us what you know so we know what words you will understand

Here is a hint.

At any point p where differentiable functions f(x) and g(x) are tangent, f(p) = g(p) and f’(p) = g’(p).
 
Jeff1801 said:
This question is new to me so I still need to learn how to do it. I don't know how to go about getting the value for k.
Click to expand...
The question is new to us, too, It's up to you to provide enough information so we know where you're coming from.

How do you think you should proceed? What are the requirements for "is tangent to"? Does it mean they must intersect? Are you SURE there is such a place where they are tangent?
 
One graph is tangent to another at a given point if they both pass through that point and have the same tangent line at that point (which means they have the same derivative there).

One graph is given by \(\displaystyle y= (x- 3)(1- x)= -x^2+ 4x- 3\) and the other by \(\displaystyle y= kx^2\). At any point they have in common, \(\displaystyle kx^2= -x^2+ 4x- 3\) so \(\displaystyle (k+1)x^2- 4x+ 3= 0\). The derivative of \(\displaystyle y= -x^2+ 4x- 3\) is \(\displaystyle y'= -2x+ 4\) and the derivative of \(\displaystyle y= kx^2\) is \(\displaystyle y'= 2kx\). Where they have the same slope they must have \(\displaystyle -2x+ 4= 2kx\) or \(\displaystyle (2k+ 2)x= 4\). From that last equation, \(\displaystyle x= \frac{4}{2k+ 2}= \frac{2}{k+1}\).

Putting that into the previous equation, \(\displaystyle (k+1)\frac{4}{(k+1)^2}- \frac{8}{k+1}+ 3= \frac{4}{k+1}- \frac{8}{k+1}+ 3= 3- \frac{4}{k+1}= 0\). \(\displaystyle \frac{4}{k+1}= 3\). \(\displaystyle 4= 3(k+ 1)\), \(\displaystyle k+ 1= \frac{4}{3}\). \(\displaystyle k= \frac{4}{3}- 1= \frac{1}{3}\).
 
And then show (or explain) that k=-1 is not a solution. (The only case that has not been dealt with in this work).
 
Yes indeed,[MATH][/MATH] you showed \(\displaystyle x= \frac{2}{k+1}\) when [MATH]k≠-1[/MATH].
 
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