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Mathematics 17 Online
OpenStudy (precal):

The graph of x-2y=9 is parallel to the normal line to the graph of f(x) when x=5. What is the value of f'(5)? Justify your answer.

OpenStudy (precal):

x-2y=9 slope is 1/2 slope that belongs to the normal line is -2 not sure what else to do

OpenStudy (precal):

ok I am reading this incorrectly. Maybe my slope is 1/2

OpenStudy (anonymous):

slope of \(x-2y=9\) is \(\frac{1}{2}\) so slope of normal line is \(-2\)

OpenStudy (anonymous):

oh i see you said that there is nothing else to do except say \(f'(5)=-2\) that is all

OpenStudy (mathmale):

Note: the slopes of two lines that are perpendicular are negative reciprocals, which satellite73's comment aptly illustrates.

OpenStudy (mathmale):

You say you're in precalculus, but I see you finding the derivative: f'(x). How's that?

OpenStudy (precal):

parallel to the normal line?

OpenStudy (precal):

My username is precal but I tend to study calculus for fun

OpenStudy (precal):

Calculus is not my strong point, I struggle with many calculus concepts. Solving 600 pages of calculus just for fun.

OpenStudy (mathmale):

It surely adds to your problem solving skills if you understand that the derivative of a function represents the slope of the tangent line to the curve of that function at a certain point / certain x-value.

OpenStudy (mathmale):

Let's go back to the beginning. Re-read the original question. What are you sure you are able to do correctly? What are you not so sure about?

OpenStudy (precal):

The graph of x-2y=9 is parallel to the normal line. y=(-9/2)+(1/2)x is the function and the slope is 1/2 that is parallel to the normal line of f(x). I am assuming that f(x) is some other function and not x-2y=9

OpenStudy (mathmale):

What leads you to assume that f(x)=-9/2 + (1/2)x?

OpenStudy (mathmale):

Let's suppose that we DON'T know f(x). What is the slope of the TANGENT line to the graph of this function at x?

OpenStudy (precal):

No, I am not assuming that f(x) is that. I solved for y to determine the slope. I don't know anything about f(x) except that this line is parallel to the normal line

OpenStudy (precal):

m=-2 this slope is the slope of the tangent line

OpenStudy (precal):

because the tangent line and normal line are perpendicular

OpenStudy (mathmale):

All right. Let's back up a bit. You are dealing with a function named f(x), right? What is the derivative of f(x)?

OpenStudy (precal):

This is some type of AP Calculus AB problem, I know I am overlooking something.

OpenStudy (precal):

f'(x)

OpenStudy (mathmale):

that's right. Now, what is the slope of any line that is PERPENDICULAR to this tangent line? Write it symbolically.

OpenStudy (precal):

-1/f'(x)

OpenStudy (mathmale):

Yes, my point exactly.

OpenStudy (mathmale):

Now, what is the slope of that given line?

OpenStudy (precal):

1/2

OpenStudy (precal):

yes, both lines are parallel and therefore have the same slope

OpenStudy (mathmale):

Sorry, my sentence was 'way off. I'm going to delete it after re-writing my question.

OpenStudy (mathmale):

We are told that the given line and the normal line to the curve at x=5 are parallel. What does that say ab out their slopes?

OpenStudy (precal):

so is my answer -1/-2 or 1/2

OpenStudy (precal):

so satellite was incorrect? Not even sure I am correct but I am following your logic

OpenStudy (precal):

other slope is negative reciprocal

OpenStudy (precal):

now I am not sure, is the solution -2 or 1/2

OpenStudy (precal):

slope is 1/2 in the beginning

OpenStudy (precal):

slope of x-2y=9 is 1/2

OpenStudy (mathmale):

I agree that the slope of the given line is 1/2. You correctly found an expression for the slope of the normal line. For review, type it out again here. Now equate those two quantities and solve for f'(5).

zepdrix (zepdrix):

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