Asked by Ifi
                Let f(x) be the parabola -x^2+16x-16. find the point (x,y) on f such that tangent line to f at (x,y) passes through the origin.
My ans is (8,48). First i do f'x then i find x and lastly i find y. I do this because the max/min point is the point where tangent line pass through. Pls correct me if im wrong.
            
        My ans is (8,48). First i do f'x then i find x and lastly i find y. I do this because the max/min point is the point where tangent line pass through. Pls correct me if im wrong.
Answers
                    Answered by
            Steve
            
    No, (8,48) is the max, so the tangent there is horizontal. It will not pass through the origin.
f'(x) = -2x+16
you want to find (h,k) where the line joining (0,0) and (h,k) has slope -2h+16
k/h = -2h+16
But, k = -h^2+16h-16, so
-h^2+16h-16 = h(-2h+16)
-h^2 + 16h - 16 = -2h^2 + 16h
h^2 - 16 = 0
h = 4,-4, so k = 32,-96
The points are thus (-4,-96) and (4,32)
Check:
f'(4) = 8 = 32/4
f'(-4) = 24 = 96/4
    
f'(x) = -2x+16
you want to find (h,k) where the line joining (0,0) and (h,k) has slope -2h+16
k/h = -2h+16
But, k = -h^2+16h-16, so
-h^2+16h-16 = h(-2h+16)
-h^2 + 16h - 16 = -2h^2 + 16h
h^2 - 16 = 0
h = 4,-4, so k = 32,-96
The points are thus (-4,-96) and (4,32)
Check:
f'(4) = 8 = 32/4
f'(-4) = 24 = 96/4
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.