Precalc0201to0204

# Precalc0201to0204 - 2.01 CHAPTER 2: POLYNOMIAL AND RATIONAL...

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2.01 CHAPTER 2: POLYNOMIAL AND RATIONAL FUNCTIONS SECTION 2.1: QUADRATIC FUNCTIONS (AND PARABOLAS) PART A: BASICS If a , b , and c are real numbers, then the graph of f x ( ) = y = ax 2 + bx + c is a parabola, provided a 0 . If a > 0 , it opens upward . If a < 0 , it opens downward . Examples The graph of y = x 2 4 x + 5 (with a = 1 > 0 ) is on the left. The graph of y = x 2 + 4 x 3 (with a = 1 < 0 ) is on the right.

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2.02 PART B: FINDING THE VERTEX AND THE AXIS OF SYMMETRY (METHOD 1) The vertex of the parabola [with equation] y = ax 2 + bx + c is h , k ( ) , where: x -coordinate = h = b 2 a , and y -coordinate = k = f h ( ) . The axis of symmetry, which is the vertical line containing the vertex, has equation x = h . (Does the formula for h look familiar? We will discuss this later.) Example Find the vertex of the parabola y = x 2 6 x + 5 . What is its axis of symmetry? Solution The vertex is h , k ( ) , where: h = b 2 a = 6 2 1 ( ) = 3 , and k = f 3 ( ) = 3 ( ) 2 6 3 ( ) + 5 = 4 The vertex is 3, 4 ( ) . The axis of symmetry has equation x = 3 . Since a = 1 > 0 , we know the parabola opens upward . Together with the vertex, we can do a basic sketch of the parabola.
2.03 PART C: FINDING MORE POINTS “Same” Example: y = x 2 6 x + 5 , or f x ( ) = x 2 6 x + 5 Find the y -intercept. Plug in 0 for x . Solve for y . In other words, find f 0 ( ) . The y -intercept is 5 . (If the parabola is given by y = ax 2 + bx + c , then c , the constant term, is the y -intercept. Remember that b was the y -intercept for the line given by y = mx + b .) Find the x -intercept(s), if any. Plug in 0 for y . Solve for x (only take real solutions). In other words, find the real zeros of f x ( ) = x 2 6 x + 5 . 0 = x 2 6 x + 5 0 = x 5 ( ) x 1 ( ) x = 5 or x = 1 The x -intercepts are 1 and 5 . Note: Observe that f x ( ) = x 2 + 1 has no real zeros and, therefore, has no x - intercepts on its graph. You can find other points using the Point-Plotting Method (Notes 1.27) . Symmetry helps!

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2.04 PART D: PARABOLAS AND SYMMETRY “Same” Example Sketch the graph of y = x 2 6 x + 5 , or f x ( ) = x 2 6 x + 5 . Clearly indicate: 1) The vertex 2) Which way the parabola opens 3) Any intercepts Make sure your parabola is symmetric about the axis of symmetry. Solution Point A is the vertex. Points B and C are the x -intercepts. Point D is the y -intercept. We get the “bonus” Point E by exploiting symmetry.
2.05 Observe that the x -intercepts are symmetric about the axis of symmetry. This makes sense, because the zeros of f are given by the QF: x = b ± b 2 4 ac 2 a The average of these zeros is b 2 a , which is the x -coordinate for the vertex and the axis of symmetry. Technical Note: If the zeros of a quadratic f x ( ) are imaginary, then they must have the same real part, which will be b 2 a . The axis of symmetry will still be x = b 2 a . There will be no x -intercepts, however. Technical Note: The x -intercept on the right (if any) does not always correspond to the “+” case in the QF. Remember, a can be negative.

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