<p> 2 Graphing Quadratics f(x) = ax + bx + c -- October 23, 2009// Slightly revised April 13, 2010</p><p>You already know: </p><p> The graph is a parabola.</p><p> It opens up if a > 0 and down if a < 0.</p><p> The y-intercept is (0,c).</p><p> The graph is symmetric about a vertical line through the vertex.</p><p>You may or may not know this: b FACT: The vertex is at x= -, y = f ( x ) V2a V V</p><p>Example 1: y = 2x2 – 12x + 5.</p><p>Opens up.</p><p>Vertex at x = –(-12)/4 = 3; y = 2*32 –12*3 + 5 = 18 – 36 + 5 = -13; vertex (3, -13).</p><p>Intercept is (0, 5). </p><p>New idea: Symmetric point to the intercept is (6, 5) -- reflect (0,5) through the axis of symmetry, x = 3.</p><p>We have three points: (3, -13), (0, 5), and (6, 5), and we know the parabola opens up. Sketch the curve by plotting the three points and drawing the parabola through them.</p><p>Here’s what we do:</p><p>1. Note whether it opens up or down. b 2. Find the vertex x= -, y = f ( x ) . V2a V V 3. Plot the intercept, and plot the symmetric point.</p><p>4. Sketch.</p><p>Example 2. y = –x2 + 10x –14</p><p>1. Opens down.</p><p>2 2. The vertex is at xV = -(10)/2(-1) = 5, yV = –(5 ) + 10*5 + 14 = 25 – 14 = 11. So, the vertex is at (5, 11). 3. Intercept (0, -14), and symmetric point (10, -14) -- reflected across the line x = 5.</p><p>4. Three points, (5, 11), (0, -14), (10, -14). Opens down. Sketch.</p><p>Example 3. y = 3x2 + 5x + 4 (Numbers not nice, but we can handle them.)</p><p>1. Opens up.</p><p>2. Vertex x = -5/(2*3) = -5/6; y = 3(-5/6)2 + 5(-5/6) + 4 = 25/12 – 25/6 + 24/6 = 23/12, a little less than 2. Vertex (-5/6, 2– ).</p><p>3. Intercept (0, 4); symmetric point (-5/3, 4) , because 2*(-5/6) = -5/3.</p><p>4. Three points, opens up; sketch.</p><p>In some special cases, other ideas make it easier:</p><p>Example 4 (Factored; or factors easily). y = (x – 3)(x + 9). </p><p>Since this is factored, the vertex must be half-way between the zeroes: xV = (3 + -9)/2 = -3; yV = (-3 – 3)(-3 + 9) = -36; Vertex (-3, -36).</p><p>Two more points are the zeroes, (3, 0) and (-9, 0). </p><p>Three points, opens up, sketch.</p><p>Example 5 (Easy to CTS): y = x2 – 10x + 1</p><p>We complete the square, but now adding and subtracting the same number on the right-hand side: y = x2 – 10x + 25 –25 + 1 = (x – 5)2 – 25 + 1 = (x – 5)2 – 24. Vertex at (5, -24). </p><p>As done previously, the y-intercept and its symmetric partner are (0, 1), (10, 1). </p><p>Three points, so sketch. (No fuss, no muss, no bother . . . )</p><p>The book uses CTS as the primary method – see the book’s Examples 1, 2, 3 in Section 2.4. The book’s Example 4 is done using the formula, like Examples 1, 2, 3 on this handout.</p><p>T. Wallgren – Spring 08 /Spring 2010</p>