Equation of the Normal Line...1

[nycmathguy]

For the function given below determine the equation of the normal line at x = 0.

f(x) = x^2+3x+1 at x = 0

1. What is a normal line?

2. Can this question be answered without taking the derivative?

Thank you.

 

[MathLover1]

f(x) = x^2+3x+1 at x = 0

1. What is a normal line?
a normal line is a line perpendicular on tangent line

2. Can this question be answered without taking the derivative?
no, to find a slope of tangent line you need derivative

 

[nycmathguy]

f(x) = x^2+3x+1 at x = 0

1. What is a normal line?
a normal line is a line perpendicular on tangent line

2. Can this question be answered without taking the derivative?
no, to find a slope of tangent line you need derivative

Can you solve this for me?

Let me see.

f ' (x) = 2x + 3

Yes?

What's next?

 

[MathLover1]

f ' (x) = 2x + 3
find a slope using given x=0

f ' (x) = 2*0 + 3=3-> slope m=3

go back to given equation, substitute x=0

y=0^2+3*0+1=1

so, we have a point (0,1) which is a point of tangency

tangent line is:
y-y1 =m(x-x1)
y-1 =3(x-0)
y-1 =3x
y =3x+1

then, a slope of normal line is m=-1/3

use slope point formula to find equation of a normal line with a slope m=-1/3 and passes through the point (0,1)

y-y1 =m(x-x1)
y-1 =-(1/3)(x-0)
y-1 =-(1/3)x
y =-(1/3)x+1

 

[nycmathguy]

f ' (x) = 2x + 3
find a slope using given x=0

f ' (x) = 2*0 + 3=3-> slope m=3

go back to given equation, substitute x=0

y=0^2+3*0+1=1

so, we have a point (0,1) which is a point of tangency

tangent line is:
y-y1 =m(x-x1)
y-1 =3(x-0)
y-1 =3x
y =3x+1

then, a slope of normal line is m=-1/3

use slope point formula to find equation of a normal line with a slope m=-1/3 and passes through the point (0,1)

y-y1 =m(x-x1)
y-1 =-(1/3)(x-0)
y-1 =-(1/3)x
y =-(1/3)x+1

Very good. I will post a few more with work shown when time allows. I may be moving soon.