Find equations of all tangent lines to the graph y=4x3−19x+40 which pass through the point (2,2) ​(not on the​ curve).

Answer 1

y=4x^3−19x+40
y' = 12x^2-19
So, at any point (x,y) the slope is 12x^2-19

Using the point-slope form

y-2 = y'(x-2)
4x^3−19x+40-2 = (12x^2-19)(x-2)
8x^2(3-x) = 0
x = 0,3
So, the tangents through (0,40) and (3,91) pass through (2,2)

The lines are thus
y-40 = -19x
y-91 = 89(x-3)

see the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D4x%5E3%E2%88%9219x%2B40,+y%3D-19x%2B40,+y%3D89(x-3)%2B91+for+-3%3C%3Dx%3C%3D4

Answer 2

I assume you meant:
y=4x^3−19x+40
let's take a quick look at the graph
http://www.wolframalpha.com/input/?i=y%3D4x3%E2%88%9219x%2B40
looks like we will have 3 different tangents from the point A(2,2)

let the point of contact be P(a,b)
obviously b = 4a^3 - 19a + 40

slope of any tangent to the curve = 12x^2 - 19
so at the point (a,b), the slope
= 12a^2 - 19

slope of AP(done the old-fashioned grade 9 way
= (b-2)/(a-2)
The key concept is that the line AP is a normal to the tangent, that is, the slopes are negative reciprocals of each other.

(b-2)/(a-2) = -1/(12a^2 - 19)
b-2 = (2-a)/(12a^2 - 19)
4a^3 - 19a + 40 - 2 = (2-a)/(12a^2 - 19)
48a^5 - 76a^3 - 228a^3 + 361a + 456a^2 - 722 = 2-a
48a^5 - 304a^3 + 456a^2 + 362a - 724 = 0
24a^5 - 152a^3 + 228a^2 + 181a - 362 = 0
What a nasty equation!!!!
(looks like somebody just made up some numbers without regard to the algebra involved)

I will let Wolfram do this for us
http://www.wolframalpha.com/input/?i=24a%5E5+-+152a%5E3+%2B+228a%5E2+%2B+181a+-+362+%3D+0

a = -2.8435
a = -1.26031
a = 1.25942 , plus 2 imaginary roots

not done yet!!!

let a = -2.8435
b = 2.0621 , I subbed it into the original
m = 12a^2 - 19 = 78.026

equation:
y - 2.0621 = 78.026(x + 2.8435)

that is one of the three tangents, repeat the last few steps for the other values of a

I sure hope I did not make any silly arithmetic errors.

Answer 3

Just go with Steve.
I was answering a totally different question I had in my mind.