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Tangent and Secant Lines Monday, January 9, 2017 Tangent and Secant Lines • Suppose you are given some curve, in this L case a circle, and a point P. • The line L passing through the point P is L 1 called a tangent line. Q1 P • Given another point Q1 on the curve, let L1 be the line passing through the points P and Q1. • L1 is called a secant line and is at times written PQ1. Monday, January 9, 2017 Tangent and Secant Lines L 2 • Pick a point Q closer to P. L 2 • Notice that the slope of the secant line L2 is Q2 closer to the slope of L compared to the L1 slope of L1. Q1 P • In fact, as we pick points closer and closer to P, the slope of the associated secant lines will get closer and closer to the slope of L. Q: So, how can we differentiate between secant lines and tangent lines? Is it enough to say that a tangent line goes through one point, while a secant line goes through two points? A: No! Tangent and Secant Lines • L is the tangent line to P, even though it intersects the curve more than once. P • However, we can remedy L this by saying that there is a neighborhood of the point P for which L doesn’t intersect the curve at any other point within this neighborhood. Q: Is this enough? Do we have enough information now to distinguish between tangent and secant lines? A: Not quite. Tangent and Secant Lines • L1 is a secant line for the curve through the point P. L1 • We can find a neighborhood of P such P that L1 doesn’t intersect any other point of the curve within the neighborhood. What then is the real distinction between tangent lines and secant lines? Tangent and Secant Lines • If we zoom in on the point P, the curve at P starts to look more and P L more like the tangent line at P. • We can think of the tangent line as a linear approximation of the curve at P. • The notion of linear approximation is a really powerful tool in math. We will study it a bit later in the course. Warning: The tangent line to a curve at a given point Ex: Cusps may NOT exist. This is equivalent to saying that there P L is no good linear approximation of the curve at the No matter how close we given point. zoom in, L will never give a linear approximation of the curve at the point P. Tangent and Secant Lines • If we zoom in on the point P, the curve at P starts to look more and P L more like the tangent line at P. • We can think of the tangent line as a linear approximation of the curve at P. • The notion of linear approximation is a really powerful tool in math. We will study it a bit later in the course. L1 P • A secant line through the point P, on the other hand, doesn’t give a linear approximation of the curve at P. • Nonetheless, secant lines have their use! Quick Review of Lines L • Recall that the equation of a line is of the form y = mx + b, where m is the slope and b is y the y-intercept. 2 (x2,y2) y – y • Let L be a line, (x1,y1) and 2 1 (x2,y2) two points on L such that x > x . y1 2 1 (x1,y1) • The slope m = (y2 - y1)/(x2 – x1). • To find the y-intercept, plug in x2 – x1 either (x1,y1) or (x2,y2) into the equation y = mx + b and solve x x2 for b. 1 Quick Review of Lines Ex: Find the equation of the line passing through the points (5,7) and (1,8). • Since 5 > 1, the slope is given by: m = (7-8)/(5-1) = -1/4 • Thus, the equation of the line is going to be of the form: y = (-1/4)x + b • Since the point (1,8) is on the line, it must be that 8 = (-1/4)(1) + b which means that b = 33/4 • Hence, the equation of the line is given by y = (-1/4)x + 33/4 Quick Review of Lines: Class Problems. Question 1: What is the slope of the line passing through (1,1) and (2,4)? Question 2: What is the slope of the line passing through (1,1) and (a,a2), for a ≠ 1? Calculating Tangent and Secant Lines • Suppose you are given the graph of a function f and a secant line through points (a,f(a)) and (b,f(b)). f f(b) • We can calculate the equation of the secant line using the technique we reviewed in the previous slide. f(a) • The slope is given by: m=(f(b) – f(a))/(b – a) a b Calculating Tangent and Secant Lines • Suppose you are given the graph of a function f and a secant line through points (a,f(a)) and (b,f(b)). f f(a + h) • We can calculate the equation of the secant line using the technique we reviewed in the previous slide. f(a) • The slope is given by: m=(f(b) – f(a))/(b – a) a a + h • Since b = a + h, for some quantity h, we can replace the formula for the slope by m = (f(a + h) – f(a))/((a + h) – a) = (f(a + h) – f(a))/h Slope of the secant line = Calculating Tangent and Secant Lines Slope of the secant line = • If we allow the quantity h to get smaller, the point (a+h,f(a+h)) is getting closer to f(a) (a,f(a)), which in turn means that the slope of the secant line through (a+h,f(a+h)) is getting closer to the slope of the tangent a line at (a,f(a)). • Thus, the slope of the tangent line is given by taking the limit as h goes to 0. Slope of the tangent line at (a,f(a)) = Calculating Tangent and Secant Lines Ex: Calculate the equation of the tangent line of f(x)=x3 at the point (2,8). • The slope of the tangent line is given by: ! !!! !!(!) lim h→0 ! • Given that the slope is 12, we can (2 + ℎ)!−2! plug in the point (2,8) to find the y- = lim h→0 ℎ intercept 8 = 12(2) + b, and so !!!"#! !"!! !! !! b = -16 = lim h→0 ! • Thus, the equation of the line is given !"#! !"!! !! by y = 12x -16. = lim h→0 ! 2 = limh→0 (12 + 6h + h ) = 12 Calculating Tangent and Secant Lines: Class Problems Consider the function f(x) = x2 Question 3: What is the slope of the line passing through (1,1) and (1 + h,(1 + ℎ)!)? Question 4: What is the slope and equation of the tangent line at (1,1)? Question 5: What is the slope and equation of the tangent line at (a,a2)? Limits of Functions L • Given a function f, and a number a, the limit of f as x approaches a is the value, L, f(x) approaches as x goes to a. f • In this case, we write limx→a f(x) = L a Limits of Functions f(a) L Note that L doesn’t have to be the same as f(a). f a Limits of Functions • We can take limits from the left hand side and the right hand side. • Here, the limit from the left f hand side is L1. We denote this by L2 limx→a- f(x) = L1 L1 • The limit from the right hand side is L2. We denote this by limx→a+ f(x) = L2 a Note: If, as in this picture, the left hand limit and the right hand limit as x approaches a are different, then the limit of f as x approaches a does not exist. Limits of Functions • It’s possible that the limit of f as x approaches a is not a number. • In this picture limx→a f(x) = ∞ • The line x = a is an example f of a vertical asymptote. • In general, we say that f has a vertical asymptote at x = a if the limit as x approaches a a from the left, right, or both is ∞ or -∞. Limits of Functions In this picture, we have limx→a- f(x) = -∞ limx→a+ f(x) = ∞ And so f limx→a f(x) does not exist. a Limits of Functions Some Limit Laws • limx→a [f(x) + g(x)] = limx→a f(x) + limx→a g(x) • limx→a [f(x) - g(x)] = limx→a f(x) - limx→a g(x) • limx→a [f(x) · g(x)] = limx→a f(x) · limx→a g(x) !(!) limx→a f(x) • limx→a = , as long as limx→a g(x) ≠ 0 !(!) limx→a g(x) ! ! • limx→a [� � ] = [limx→a f(x)] for n ∈ ℤ Limits of Functions Limit Theorems Direct Substitution property: If f is a polynomial or a rational function defined at a, then limx→a f(x) = f(a) Theorem: If f(x) ≤ g(x) in some neighborhood of a and limx→a f(x) and limx→a g(x) exist, then limx→a f(x) ≤ limx→a g(x). g f a Limits of Functions Limit Theorems Squeeze Theorem: If f(x) ≤ h(x) ≤ g(x) in some neighborhood of a and limx→a f(x) = limx→a g(x) = L, then limx→a h(x) = L a Limits of Functions: Class Problems Question 6: Next Class • Next time, we will give a precise definition of limits.
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