Multiple tangents

The question asks,

Find all tangents to the curve

f( x)=x^2+3x+3

which go through the point

(1,-2)

This is what I've done so far,

f'( x)=2x+3

Gradient of tangest when

x=1

is,

m=2(1)+3

m=5

Equation of tangest is therefore,

y-y_1=m(x-x_1)

y-(-2)=5(x-1)

y+2=5x-5

y=5x-7

That's the equation for one of the tangents, but I understand that there's another tangent to the curve, as the point (2,1) is 'below' the curve and as a result, there's a sort of a triangular shape which the two tangents create.

How would I go about finding the other tangent?

Thanks.

(edited 13 years ago)

f'(x) is wrong

Original post by volts

f'(x) is wrong

Sorry, that was a typo. I've edited it now.

Reply 3

volts

The point (1,-2) doesn't lie on f(x). I think you assume it does.

Reply 4

hollywoodbudgie

Original post by volts

The point (1,-2) doesn't lie on f(x). I think you assume it does.

The point

(1,-2)

lies on the tangents, not on the curve

f( x)

.

So in other wording, it's two tangents to the curve

f( x)

which intersect at the point

(1,-2)

(edited 13 years ago)

Reply 5

volts

Original post by hollywoodbudgie

The point

(1,-2)

lies on the tangents, not on the curve

f( x)

.

So in other wording, it's two tangents to the curve

f( x)

which intersect at the point

(1,-2)

So why have you evaluated f'(1)?
Also, 5x-7 = f(x) has no real solution

(edited 13 years ago)

Reply 6

hollywoodbudgie

Original post by volts

So why have you evaluated f'(1)?

Because I don't know how to work this out, hence I made this thread to ask for help.

Reply 7

anshul95

Original post by hollywoodbudgie

The point

(1,-2)

lies on the tangents, not on the curve

f( x)

.

So in other wording, it's two tangents to the curve

f( x)

which intersect at the point

(1,-2)

You know that the tangent passes through (1,-2). However, you also know that at the tangent crosses the parabola. So what I would do is find the possible coordinates which lies on the parabola. To do this note that the y coordinate at x1 is x1^+3*x1+3. Now using the formula for the equation of a line with only one point given you get an expression for the possible tangents. You know that the tangents must pass through 1,-2 so substitute these values in for x and y (not x1 and y1). Solve the resulting quadratic equation and you have found the x coordinates which satisfy the conditions given in the question. I'll leave the rest to you

Reply 8

hollywoodbudgie

anshul95

note that the y coordinate at x1 is x1^+3*x1+3.

Did that equation come out right, I don't particularly understand what the bolded symbol means, '^'. Would it be possible for you to write that out a little clearer please?

Thanks.

Reply 9

nuodai

The gradient of the tangent at a point

(c, f(c))

is given by

f'(c)

. So, the equation of a tangent that goes through the point

(a,b)

and has gradient

f'(c)

y-b = f'(c)(x-a)

. But you also want the point

(c, f(c))

to lie on the line, and so you must have

f(c)-b = f'(c)(c-a)

. Solve this for

c

to find the

x

-values where the tangent of the curve goes through the point. [You have to work out what to use for a,b by the way.]

Reply 10

Physics Enemy

It's 2 tangents which happen to pass through (1, -2). Doesn't mean they touch the curve at this point. They can pass through (1, -2) yet touch the curve elsewhere.

Tangent: y = mx + c; (1, -2) satisfies this
=> -2 = m + c => c = -2 - m
=> y = mx - (m+2)

Since the tangents touch the curve:
mx - (m+2) = x^2 + 3x + 3
=> x^2 + (3 - m)x + (5 + m) = 0

Tangent => Discrim = 0
=> (3 - m)^2 - 4(m + 5) = 0
=> 9 - 6m + m^2 - 4m - 20 = 0
=> m^2 - 10m - 11 = 0
=> (m + 1)(m - 11) = 0
=> m = 11, m = -1

Sub m = 11, m = - 1 into y = mx - (m+2) to get y = 11x - 13 and y = -x - 1.

(edited 13 years ago)

Reply 11

hollywoodbudgie

Original post by Physics Enemy

Thank you

Reply 12

Physics Enemy

Original post by hollywoodbudgie

Thank you

No worries. If it's any consolation, I didn't have a clue what some of the other posters were on about either. Not sure why people make Maths harder than it has to be; it's already hard enough by itself. :smile: