8-2 Quadratic Functions (Part # 1)

Example 1
A) The vertex is (4,3) 
 B) The vertex is (-3,-3) 
 
   
   
  

Example 2 - Vertex Formula
a) y = 2x^2+4x  
 
 
 
 
 we get the a, b, c in the formula y = ax^2 + bx + c 
 
 
 a = 2 
 
 
 b = 4 
 
 
 c = 0 
 
 
 
 
 then we plug in a, b into the formula 
 
 
 x = -b / (2a) 
 
 
 x = -4 / (2*2) = -4/4 = -1 
 
 
 
 
 
 
 
 
 so the axis of symmetry is x = -1 
 
 
 to find the vertex 
 
 
 so the known vertex data is  (-1, y) 
 
 
 to find y for the vertex we need to plug -1 into the formula y = 2x^2+4x  
 
 
 y = 2*-1^2+4*-1 = -6 
 
 
 so the vertex is (-1, -6) 
 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=- 
 
 
 B) y = -x^2 + 4x - 5  
 
 
 
 
 we get the a, b, c in the formula y = ax^2 + bx + c 
 
 
 a = -1 
 
 
 b = 4 
 
 
 c = -5 
 
 
 
 
 then we plug in a, b into the formula 
 
 
 x = -b / (2a) 
 
 
 x = -4 / (2*-1) = -4/-2 = 2 
 
 
 
 
 
 
 
 
 so the axis of symmetry is x = 2 
 
 
 to find the vertex 
 
 
 so the known vertex data is  (2, y) 
 
 
 to find y for the vertex we need to plug -1 into the formula y = 2x^2+4x  
 
 
 y = -2^2 + 4 * 2 - 5 = -1 
 
 
 so the vertex is (2,  -1 )

Example 3 + up/down test
up/down test 
 
 opens upwards because a is Positive 
  opens downward because a is Negative 
 
 Example 3 
 
 a) y= x^2 + 3x + 4

 
 opens upwards because a is Positive 
 
 
 b) y = -3x^2 + 5x 
 
  opens downward because a is Negative 
 
 
 c) y = 2x - x^2 + 6 
 
  opens downward because a is Negative

Example 4 Graph f(x) = x^2 -2x - 8
Steps to Graph ax^2 + bx + c 
 
 Find the vertex and the axis of symmetry. Sketch these in. 
 Find the x-intercept by plugging in 0 for y. 
 Find the y-intercept by plugging in 0 for x. 
 Reflect your points across the axis of symmetry and connect your dots with a smooth U-shaped (not V-shaped) curve. 
 
 Graph f(x) = x^2 -2x - 8  
 
 for now, I'm just gonna type my work and figure out what to do next 
 
 find the line of symmetry -
 
 a = 1, b = -2, c = -8 
 
 
 use this to find the vertex
 
 x = (b/2a) 
 x = -(-2) / 2(1) = 1 
 
 
 since we know that the along the x axis at 1 will be the vertex we replace x with 1 in the original formula
 
 x=1

 
 y = x^2 -2x - 8 
 y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9  
 y = -9 
 
 
 the vertex is (1, -9) 
 since the vertex is -1,-9 we know that x=1 is the axis of symmetry 
 finding the y-intercept is the easiest to start with because we just replace x with 0 
 x = 0 | y = x^2 -2x - 8 
 y = 0 - 8 = -8 
 y-intercept = (0,-8) 
 so so To find the x-intercepts, you can set y equal to zero and solve for x: 
 
 y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a 
 
 
 x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1) 
 x = (2 ± sqrt(4 + 32)) / 2 
 x = (2 ± sqrt(36)) / 2 
 x = (2 ± 6) / 2 
 x = 8 / 2 or x = -4/ 2 
 x = 4 or x = -2  
sooooo (-2,0) & (4,0) 
 
 
 so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work  
 
 symmetry line = x = 1,  
 calc'd x-intercept 0,-8
 
 the symmetry line is 1 and the known point is  0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8) 
 
 
  since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
 
 y = 3^2 -3*2 - 8 =  -5 
 
 soooo the new point is (3,-5) if we mirror that along 1,-9 we get  (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate 
 
 
 
 
 so all points are:
 
 (1, -9) 
 (0,-8) 
 (2, -8) 
 (3,-5) 
 (-1,-5)

Example 5: Graph y = x^2 + 2x + 3
Find the vertex and the axis of symmetry. Sketch these in. • Find the x-intercept by plugging in 0 for y. • Find the y-intercept by plugging in 0 for x. • Reflect your points across the axis of symmetry and connect your dots with a smooth U-shaped (not V-shaped) curve. 
 
 fix the fallowing 
 
 a = 1, b = 2, c = 3 
 x^2 + 2x + 3 
 
 
 find the line of symmetry -
 
 x = (b/2a) 
 x = -(2) / 2(1) =1 
 
 
 use this to find the vertex 
 since we know that the along the x axis at -1 will be the vertex we replace x with 1 in the original formula
 
 x=1

 
 y = x^2 -2x - 8 
 y = 1^2 + -2 * 1 - 8 = 1- 2 - 8 = -9  
 y = -9 
 
 
 the vertex is (1, -9) 
 since the vertex is -1,-9 we know that x=1 is the axis of symmetry 
 finding the y-intercept is the easiest to start with because we just replace x with 0 
 x = 0 | y = x^2 -2x - 8 
 y = 0 - 8 = -8 
 y-intercept = (0,-8) 
 so so To find the x-intercepts, you can set y equal to zero and solve for x: 
 
 y = 0 | x = (-b ± sqrt(b^2 - 4ac)) / 2a 
 
 
 x = (-(-2) ± sqrt((-2)^2 - 4(1)(-8))) / 2(1) 
 x = (2 ± sqrt(4 + 32)) / 2 
 x = (2 ± sqrt(36)) / 2 
 x = (2 ± 6) / 2 
 x = 8 / 2 or x = -4/ 2 
 x = 4 or x = -2  
sooooo (-2,0) & (4,0) 
 
 
 so since we know 3 y axis points on the graph and the axis of symmetry we can get another point without doing much work  
 
 symmetry line = x = 1,  
 calc'd x-intercept 0,-8
 
 the symmetry line is 1 and the known point is  0 since 1-0 = 1 we can add that to the x coordinate of y and keep the same y coordinate to get the mirrored point making another point on the graph (2,-8) 
 
 
  since we need one more point for the graph we can choose say x=3, | x^2 -2x - 8
 
 y = 3^2 -3*2 - 8 =  -5 
 
 soooo the new point is (3,-5) if we mirror that along 1,-9 we get  (-1, -5 ) because 3 is 2 more than 1, and 2 less than 1 is -1. we also keep the same y coordinate 
 
 
 
 
 so all points are:
 
 (1, -9) 
 (0,-8) 
 (2, -8) 
 (3,-5) 
 (-1,-5)

Example 6: Graph y = 2x^2 - 8x
• Find the vertex and the axis of symmetry. Sketch these in. • Find the x-intercept by plugging in 0 for y. • Find the y-intercept by plugging in 0 for x. • Reflect your points across the axis of symmetry and connect your dots with a smooth U-shaped (not V-shaped) curve.

Example 7: h = 16t^2 + 72t + 520
Suppose a particular “star” is projected from a firework at a starting height of 520 feet with an initial upward velocity of 72 ft/sec. 
 The equation: 
 h = 16t^2 + 72t + 520 
 gives the star’s height h in feet at time t in seconds. 
 a) How long will it take for the star to reach its maximum height? 
 
 b) What is the maximum height?