Trouble understanding the relationship between f(x) and f(x+2)

[joshuamd]

This is the given question.

Given that
y= −2x+4 is tangent to y=h(x), determine a tangent line to y = h(x+2)

Not sure why they went with h for the variable as its a variable used in the limit definition equation, but here's all I've done so far

[I replaced variable 'h' with variable 't' because of the question using h(x) instead of f(x)]


Not sure where to go from here. Any help is appreciated!
Also it's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

 

[Deleted member 4993]

This is the given question.

Given that
y= −2x+4 is tangent to y=h(x), determine a tangent line to y = h(x+2)

Not sure why they went with h for the variable as its a variable used in the limit definition equation, but here's all I've done so far

[I replaced variable 'h' with variable 't' because of the question using h(x) instead of f(x)]

View attachment 28768
Not sure where to go from here. Any help is appreciated!
Also it's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

Given that
y= −2x+4 is tangent to y=h(x), determine a tangent line to y = h(x+2)

Please post the COMPLETE and EXACT problem

- as posted, the problem statement seem incomplete.​

 

[joshuamd]

Please post the COMPLETE and EXACT problem

Given that
y= −2x+4 is tangent to y=h(x), determine a tangent line to y=h(x+2)
(Hint: what is the relationship between the graph ofy=h(x)andy=h(x+2)?)

Here is the complete and exact problem as per my study guide.

 

[Dr.Peterson]

This is the given question.

Given that
y= −2x+4 is tangent to y=h(x), determine a tangent line to y = h(x+2)

Not sure why they went with h for the variable as its a variable used in the limit definition equation, but here's all I've done so far

[I replaced variable 'h' with variable 't' because of the question using h(x) instead of f(x)]

View attachment 28768
Not sure where to go from here. Any help is appreciated!
Also it's worth mentioning that I'm beginning Calc 1 so I can't use L'Lopital's Rule, power rule, quotient rule, or the chain rule unfortunately.

This is a lot easier than you are thinking. You don't know where the line is tangent to h, and you are free to find any tangent line to h(x+2).

The fact that they used the letter h, which is used in one form of the definition of the derivative, doesn't mean this problem has anything to do with derivatives! There is no limit as h approaches zero here.

What happens to the graph of a function when you replace x with x+2?

Do the same thing to the given line.

 

[joshuamd]

This is a lot easier than you are thinking. You don't know where the line is tangent to h, and you are free to find any tangent line to h(x+2).

The fact that they used the letter h, which is used in one form of the definition of the derivative, doesn't mean this problem has anything to do with derivatives! There is no limit as h approaches zero here.

What happens to the graph of a function when you replace x with x+2?

Do the same thing to the given line.

Oh that is way easier than I thought. Thank you. So once I plug in x+2 (ang get -2), I just need to use the definition of a derivative equation, plug in 0 for h, and use that slope value to form a slope equation for the tangent line? Or is it just y=-2x?

 

[Dr.Peterson]

Thank you. So once I plug in x+2 (ang get -2), I just need to use the definition of a derivative equation, plug in 0 for h, and use that slope value to form a slope equation for the tangent line?

No. You don't need a derivative, and definitely shouldn't set the function h to 0.

Or is it just y=-2x?

Yes. You've shifted y = -2x + 4 left 2 units, by replacing x with x+2, resulting in y = -2(x+2) + 4 = -2x. Since the unknown function h was shifted the same way, the new line is tangent to the new curve. There are, of course, many other tangents to it, but this is the only one we know.

 

[joshuamd]

No. You don't need a derivative, and definitely shouldn't set the function h to 0.

Yes. You've shifted y = -2x + 4 left 2 units, by replacing x with x+2, resulting in y = -2(x+2) + 4 = -2x. Since the unknown function h was shifted the same way, the new line is tangent to the new curve. There are, of course, many other tangents to it, but this is the only one we know.

Ok thank you again for the help. Sorry if my questions are dumb, but I do really appreciate the response.