x^2 + 5, x > 3 f (x) = x-7, x < 3 x^2, x = 3 f(12) = 149 f (0) = -7 f (3) = 9 Remember what a function is: for every valid x-value, there is only one valid y-value. 4x + 7y = 12 <== is this a function? Is y a function of x? 7y = 12 - 4x y = (12-4x)/7 <== No matter what x-value I plug in, I get only y-value back, so it is a function. 7x + y^2 = 42 y^2 = 42 - 7x y = +/- sqrt (42-7x) So, it is possible for y to have more than one value depending on what x is. Rule of thumb: Based on the mathematics we have seen thus far, if there's a y^2 in there somewhere it's probably not a function. If there's no y^2, it probably is one. (But this is a guideline not a rule. You need to solve for y anyway.) Another way to test for functions is to graph them and apply the "vertical test". Draw the graph, and see if it possible to draw a vertical (up-and-down) line that intersects the graph twice (or more). If it is possible to do this, then it is not a function. If it is not possible to do this, it is a function. LINES The distance function between any two points on a line: sqrt ((x2-x1)^2 + (y2-y1)^2 ) : a form of the Pythagorean Theorem (0,5) (2,-3): What is the distance between these two? sqrt ( (0-2)^2 + (5--3)^2 ) = sqrt ( (-2)^2 + 8^2 ) = sqrt (4+64) = sqrt (68) = sqrt (4 x 17) = sqrt (4) x sqrt (17) = 2 sqrt (17) What point on the line y = 2x+1 is exactly 5 units away from the point (1,1)? Suppose (x,y) is some point on that line. sqrt ( (x-1)^2 + (y-1)^2 ) = 5 Square both sides (x-1)^2 + (y-1)^2 = 25 (x-1)^2 + (2x+1-1)^2 = 25 (x-1)^2 + (2x)^2 = 25 FOIL:(x-1)(x-1) x^2 -x -x +1 = x^2 -2x+1 x^2 - 2x + 1 + 4x^2 = 25 5x^2 - 2x+1 = 25 -25 -25 5x^2 - 2x - 24 = 0 b^2 - 4ac = (-2)^2 - 4x5x(-24) 4 + 480 = 484 = 22^2 <== This can factor ac = -120 b = -2 10 -12 /5 /5 -2 12/5, so these are the possible x-values. y = 2x+1: If x=-2, y=-3 If x=12/5, y = 24/5 + 5/5 , 29/5 (-2,-3) (12/5,29/5) are the two points on y=2x+1 that are exactly 5 units away from (1,1).