Matematics 5 Workseet 11 Geometry, Tangency, and te Derivative Problem 1. Find te equation of a line wit slope m tat intersects te point (3, 9). Solution. Te equation for a line passing troug a point (x 0,y 0 ) wit slope m can be written in slope-intercept form as y y 0 = m(x x 0 ). Tus a line wit slope m passing troug (3, 9) can be given by te equation y 9=m(x 3). Problem 2. Define te function f by f(x) =x 2. Wat must m be so tat te line you found in Problem 1 intersects te grap of f at only te point (3, 9) and at no oter point? Grap te function f and te line you ave found. Solution. Te grap of f will intersect te line from Problem 1 at any point (x, y) suc tat bot te equation defining te line and te equation defining te grap of f are satisfied. Tat is, te points of intersection are te solutions to te system of equations { y = x 2, y 9=m(x 3). Keep in mind tat we are not very interested in wat te solutions are we are primarily concerned wit knowing ow many solutions tere are. Specifically, we want to find values of m so tat tere is exactly one solution. By substitution, we ave x 2 9=m(x 3) x 2 mx (9 3m) =0. Tis equation is quadratic in x, ence tere will be only one possible value for x if and only if te discriminant,, is zero. Recall tat te discriminant of a quadratic function g(x) =ax 2 + bx + c is given by =b 2 4ac. In tis case, te discriminant is given by =m 2 +4(9 3m) =m 2 12m +36. Setting te discriminant to zero and solving for m, weobtain 0=m 2 12m +36=(m 6) 2 = m =6. 1
20 y = f(x) 15 10 (3, 9) 5 y =6x 9 2 4 Figure 1: Te grap of f and te line wit equation y =6x 9, wic intersects f at exactly one point. Terefore a line intersecting te grap of f at only te point (3, 9) must ave slope m = 6, i.e. it must ave equation y 9=6(x 3) = y =6x 9. Te grap is given in figure 1. Problem 3. Grap te line L tat intersects te grap of f (given in Problem 2) at te point (3, 9) and at (3 +, (3 + ) 2 ). Define m() to be te slope of tis line wen is not zero. Calculate m() and ten calculate m(). 0 How does tis compare to te slope m tat you found in Problem 2. Solution. Te grap will look someting like wat is sown in figure 2. Recall tat te slope of a line passing troug two points (x 1,y 2 )and (x 2,y 2 )isgivenby m = y 1 y 2. x 1 x 2 Tus, as L passes troug te points indicated in te figure, we ave tat te slope of L (assuming tat 0)isgivenby m() = (3 + )2 9 (3 + ) 3 = 9+6 + 2 9 = 2 +6 = +6. 2
20 y = f(x) 15 L 10 (3, 9) 5 (3 +, (3 + ) 2 ) 2 4 Figure 2: Te grap of f and te line L, wic passes troug te points (3, 9) and (3 +, (3 + ) 2 ). Note tat te picture will look different, depending on te coice of. Here, we ave cosen to be someting negative. Computing te it at =0,weave m() = +6=6, 0 wic is te slope of te line we got in problem 2. Problem 4. Repeat Problems 1 3, but tis time let te point be (a, a 2 ). Solution. We will not give te graps again, as te previous figures give us most of te information tat we require. Te equation of a line passing troug (a, a 2 )isgivenby y a 2 = m(x a) = y = mx ma + a 2, were m is te slope of te line. In order to determine a value of m suc tat tis line intersects te grap of f only at te point (a, a 2 ), we need to find m suc tat te system { y = x 2, y = mx ma + a 2 3
as only one solution. By substitution, want to find an m suc tat x 2 mx +(ma a 2 )=0 as only one solution. As above, tis will occur only wen te discriminant is zero, tat is, wen 0=m 2 4(ma a 2 )=m 2 4am +4a 2 =(m 2a) 2. Tis implies tat m =2a. Terefore, summarizing te desired results from problems 1 and 2, we conclude tat te line wit equation y =(2a)x (2a)a + a 2 =2ax a 2 will intersect te grap of f te point (a, a 2 ), and at no oter point. Repeating te setup for problem 3, take L to be te line passing troug te point (a, a 2 )andtepoint(a +, (a + ) 2 ). Te slope of tis line is given by m() = (a + )2 a 2 (a + ) a = a2 +2a + 2 a 2 = 2 +2a = +2a. Taking te it at =0,weave m() = +2a =2a, 0 wic is te slope of te line we described above. Problem 5. Let g(x) =x 3.Now,set g(2 + ) g(2) m =. Find te equation of te line L wit slope m tat intersects (2, 8). Now calculate Calculate Now calculate g(x) L(x). (g(x) L(x)). Wat does tis tell you about te line L? g(x) L(x). x 2 4
Solution. First, computing m, weobtain g(2 + ) g(2) m = (2 + ) 3 2 3 = 2 3 +3 2 2 +3 2 2 + 3 2 3 = 3 +6 2 +12 = = 2 +6 +12 0 =12. An equation for L is given by y 8=m(x 2) = 12(x 2) = 12x 16, wic implies tat y = 4x. Setting L(x) := 4x, we proceed wit te computations: g(x) L(x) =x 3 12x +16, wic implies tat (g(x) L(x)) = (x3 12x+16) = 2 3 12(2)+16 = 8 24+16 = 0. Lastly, we ave g(x) L(x) x 3 12x +16 = x 2 x 2 = x 2 +2x 8 (by polynomial long division) =2 2 +2(2) 8 =4+4 8 =0. Now... wat does tis tell us about L? Tis is actually a rater subtle question. Tinking in terms of elementary transformations, te expression g(x) L(x) is te grap of g, but sifted down by te line L. Te end result is sown in figure 3. If you would like to see someting sligtly more intuitive, I ave created an animation, wic is available on my website, via te link ttps://goo.gl/uaeuss We know tat g(x) L(x) is a polynomial in fact, as we noted above, it is te polynomial g(x) L(x) =x 3 12x +16. 5
100 75 50 25 4 2 25 2 4 50 75 100 Figure 3: Te grap of g(x) =x 3 is sown in red, te line L is sown in blue, and te grap of te function g(x) L(x) is sown wit a eavy black line. Tis polynomial as a zero at x =2(sinceg(2) = L(2)). To understand te beaviour of te grap at tis point, our usual tecnique is to completely factor te polynomial (assuming tat tis is possible), ten caracterize beaviour at a zero based on te multiplicity of tat zero. In tis case, we can factor to get g(x) L(x) =(x +4)(x 2) 2 (te first step of te factorization can be done via syntetic or polynomial long division; after tat, te result is quadratic and te factorization muc less difficult). We ten conclude tat te grap of g(x) L(x) is tangent to te x-axis at x = 2 (i.e. te grap bounces off te axis at tat point). Te ultimate conclusion is tat te grap of g(x) L(x) is tangent to te x-axis at x = 2, wic implies tat te graps of g and L are tangent to eac oter at x = 2 (tink about playing te animation backwards). So wy are we taking a it of te form g(x) L(x)? x 2 Imagine tat we ave a polynomial p(x) wit a root at x = a. Witout factoring p (because, in reality, it is generally almost impossible to factor 6
polynomials), ow can we tell te multiplicity of te root? One idea is to factor out te root, and see wat we get. Tat is, we consider te polynomial p(x) x a. If x a p(x) x a =0, tis implies tat x = a is a root of p(x)/(x a), wic furter implies tat it is a root p wit at least multiplicity 2. Hence we can use its to determine someting about te multiplicity of roots. In te context of te current problem, we know tat g(x) L(x) asa root at x = 2, and te fact tat g(x) L(x) =0 x 2 tells us tat tis is a root of at least multiplicity 2. Since it is a root of at least multiplicity 2, te grap of g(x) L(x) will be tangent to te x-axis at x = 2, wic again implies our ultimate conclusion tat L(x) is (in some sense) tangent to g(x) at te point (2, 8). 7