# Thread: equation of tangent, normal line and vertex?

1. ## equation of tangent, normal line and vertex?

for the function f(x) = 2x^2 - 8x - 5
how do i find equation of tangent, and equation of normal line to the curve y = f(x) at (1, f(1)) ? and vertex?

as far as i know, for the normal line,
f(#), so it would be (1,f(#))
and - 1/f'(x) is normal line
but i don't know how to find "equation" of normal line, and equation of tangent. and vertex........?

2. Recall the point slope form of a line through(x0,y0)

y = m(x-x0) + y0

For the tangent line m = f '(x0) for the normal line m =-1/f ' (x0)

For the vertex don't forget your college algebra --complete the square!!

3. so from here, y = m(x-x0) + y0, is m = f'(x) ?

4. m = f ' (1) since you want the tangent line at x = 1

Note if you use f ' (x) you don't even get a linear equation

5. Originally Posted by Calculus26
Recall the point slope form of a line through(x0,y0)

y = m(x-x0) + y0

For the tangent line m = f '(x0) for the normal line m =-1/f ' (x0)

For the vertex don't forget your college algebra --complete the square!!
i found equation of tangent and normal line, however, i still have no clue how to find vertex using calculous..

6. Hint the vertex is the minimum---What is f ' at a minimum ?

7. Complete the square to answer Caculus26's question.

8. Originally Posted by Calculus26
Hint the vertex is the minimum---What is f ' at a minimum ?
i've tried this way (if this wasn't where you were trying to get me on)

so im looking for the vertex... which will have horizontal line,
i.e. f'(x) = 0

f'(x) = 4x - 8 = 0
x = 2

v= (2, f(2))

f(2) = 2(2^2)-8(2)-5
= -13

so vertex will be (2,-13)

did i do it correctly?

9. you got it now