Calculus Problem

[Wheatmaster]

bah i screwed on the test and i want to know how to do this problem:

A tangent line of f(x)=x^2+2 has a y-intercept of -1.

A) Find the coordinates of the point of tangency.

B) Find the equation of the tangent line.

 

[us3rnotfound]

A)
B) :beer:

 

[AgaBoogaBoo]

Originally posted by: us3rnotfound
A)
B) :beer:

lol

 

[suklee]

A)
B) :beer:

 

[silverpig]

f`x = 2x

y = mx + b
x^2 + 2 = 2x(x) - 1
x^2 + 3= 2x^2
3 = x^2
x = Sqrt(3)

y = Sqrt(3)^2 + 2
y = 5

Point of tangency (x,y) = (Sqrt(3),5)

Equn of tgt line:

y - 5 = 2(Sqrt(3)) (x - Sqrt(3))
y = 2(Sqrt(3))x - 6 + 5
y = 2(Sqrt(3)) x - 1

I hope I didn't mess it up.

 

[suklee]

.

 

[Soccer55]

f'(x) gives the slope of the line tangent to f(x) at x. Since you're looking for the tangent line, you know it will have the form y=mx+b. m=f'(x) and b=-1 from what is given. Then you find where mx+b and f(x) intersect to get the x coordinate of the point of tangency. Plug that x coordinate into f(x) to get the y coordinate. Then you're done with A). To get B), put the x coordinate of the point of tangency into f'(x) so that when you do y=mx+b, you get the equation of the tangent line.

-Tom

EDIT: Bah, beaten to it. silverpig's solution is the way you do what I described above.

 

[Wheatmaster]

Originally posted by: silverpig
f`x = 2x

y = mx + b
x^2 + 2 = 2x(x) - 1
x^2 + 3= 2x^2
3 = x^2
x = Sqrt(3)

y = Sqrt(3)^2 + 2
y = 5

Point of tangency (x,y) = (Sqrt(3),5)

Equn of tgt line:

y - 5 = 2(Sqrt(3)) (x - Sqrt(3))
y = 2(Sqrt(3))x - 6 + 5
y = 2(Sqrt(3)) x - 1

I hope I didn't mess it up.

for part B, why is the slope 2sqrt3 and not just 2x?

 

[silverpig]

Originally posted by: Wheatmaster

Originally posted by: silverpig
f`x = 2x

y = mx + b
x^2 + 2 = 2x(x) - 1
x^2 + 3= 2x^2
3 = x^2
x = Sqrt(3)

y = Sqrt(3)^2 + 2
y = 5

Point of tangency (x,y) = (Sqrt(3),5)

Equn of tgt line:

y - 5 = 2(Sqrt(3)) (x - Sqrt(3))
y = 2(Sqrt(3))x - 6 + 5
y = 2(Sqrt(3)) x - 1

I hope I didn't mess it up.

for part B, why is the slope 2sqrt3 and not just 2x?

It is 2x... with x = Sqrt(3). You want the slope at that point.

Basically f`x = 2x means that the slope of the tangent to your cuve is 2 times whatever the x coordinate is. We want the slope at the point of tangency for this line, which is when x = Sqrt(3)

 

[Muzzan]

Originally posted by: silverpig
f`x = 2x

y = mx + b
x^2 + 2 = 2x(x) - 1
x^2 + 3= 2x^2
3 = x^2
x = Sqrt(3)

y = Sqrt(3)^2 + 2
y = 5

Point of tangency (x,y) = (Sqrt(3),5)

Equn of tgt line:

y - 5 = 2(Sqrt(3)) (x - Sqrt(3))
y = 2(Sqrt(3))x - 6 + 5
y = 2(Sqrt(3)) x - 1

I hope I didn't mess it up.

If x^2 = 3, then x is not sqrt(3), it could be -sqrt(3) as well. (Not that you have to make any significant changes to your solution to take this into account).

 

[chuckywang]

Originally posted by: Muzzan

Originally posted by: silverpig
f`x = 2x

y = mx + b
x^2 + 2 = 2x(x) - 1
x^2 + 3= 2x^2
3 = x^2
x = Sqrt(3)

y = Sqrt(3)^2 + 2
y = 5

Point of tangency (x,y) = (Sqrt(3),5)

Equn of tgt line:

y - 5 = 2(Sqrt(3)) (x - Sqrt(3))
y = 2(Sqrt(3))x - 6 + 5
y = 2(Sqrt(3)) x - 1

I hope I didn't mess it up.

If x^2 = 3, then x is not sqrt(3), it could be -sqrt(3) as well. (Not that you have to make any significant changes to your solution to take this into account).

Right, so the other solution is y = -2*sqrt(3) x - 1