Find equations of all tangent lines to the graph y=4x3−19x+40 which pass through the point (2,2) (not on the curve).
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
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.
Just go with Steve.
I was answering a totally different question I had in my mind.
To find the equations of tangent lines to the graph that pass through the point (2,2), we need to first find the values of x where the tangent lines intersect the curve. Then, we can calculate the slope of the tangent lines at those intersection points and use the point-slope form of a line equation to obtain the equations of the tangent lines.
Step 1: Finding the intersection points
- Let's denote the point of intersection as (x₁, y₁) on the curve y = 4x^3 - 19x + 40.
- Since the tangent line passes through the point (2,2), we have x₁ = 2 and y₁ = 2.
- Substituting these values into the equation, we get 2 = 4(2)^3 - 19(2) + 40.
- Simplifying, we find 2 = 32 - 38 + 40, which gives us 2 = 34.
- This indicates that there are no intersection points between the tangent lines and the curve at the point (2,2).
Since there are no intersection points, we cannot determine the equations of tangent lines passing through the point (2,2).