Quadratic Functions and their graphs Lesson 1.7 A quadratic function- f(x) = ax2 + bx + c where a=/ 0 The graph of a quadratic function is called a parabola Two very special parts of a parabola: Vertex: The ‘turning’ point It is either a maximum or minimum. Axis of symmetry: A vertical Line that passes through the Vertex.

The axis of symmetry can always be found by calculating: x = - (b) 2a Vertex: ( - b , f( - b) ) 2a 2a Discriminant: If b2 – 4ac > 0 Parabola crosses x-axis twice. There will be two x-intercepts. b2 – 4ac = 0 Parabola is ‘tangent’ to x-axis. There is only one x-intercept. b2 – 4ac < 0 Parabola never crosses the x-axis. No x-intercepts.

The axis of symmetry is a vertical line midway between the x-intercepts. Therefore it is the ‘average’ of the x-intercepts. Example: Find the intercepts , the axis of symmetry , and the vertex , of this parabola. y = (x+4)(2x – 3) To find x-intercepts: Replace y with 0. 0 = (x+4)(2x – 3) Set each factor = 0 x + 4 = 0 & 2x – 3 = 0 x = - 4 x = 3/2 Since the x-intercepts have already been found Find the average of these to find the axis. x = -4 + 3/2 = -2.5 = -1.25 2 2 Now  Vertex = (-1.25, f(-1.25)) (-1.25,-15.125) Sketch the graph

Example: Sketch the graph of the parabola. Label the intercepts, the axis of symmetry, and the vertex. y = 2x2 – 8x + 5 1st find the Axis of Symmetry X = -(-8) = 8 = 2 2(2) 4 Now find the vertex: V = (2, f(2)) = (2,- 3) On this one, since b2 – 4ac = 44 , Which is not a perfect square, this can- Not be factored  use the quadratic Formula to find x intercepts. X = 0.78 & 3.22 If you remember ‘c’ is always the y-intercept Sooo  y-int = 5. Draw the ‘vertical axis’ at x = 2, Plot the vertex at (2,-3) Estimate the x-intercepts at 0.78 and 3.33, Plot the y-intercept at y = 5 and plot its symmetry point at (4,5) and sketch the parabola! If the equation can be written in the form of : y = a(x – h)2 + k --- vertex -- (h,k) axis of sym. -- x = ‘h’

If the equation can be written in the form of : y = a(x – h)2 + k --- vertex -- (h,k) axis of sym. -- x = ‘h’ Example: a) Find the vertex of the parabola y = - 2x2 + 12x + 4 by completing the square. y = -2x2 + 12x + 4 = 0 1st subtract 4 from both sides y – 4 = -2x2 + 12x factor out a – 2 from both ‘x’terms y – 4 = -2(x2 – 6x ) complete the square inside the parentheses b = -6/2 = (-3)2 = 9  now add 9 inside the ( ) but add -2(9) or -18 to other side y – 4 – 18 = -2(x2 -6x +9)  change to this look y – 22 = -2(x – 3)2 Now add 22 back to the right side y = -2(x – 3)2 + 22 line up y = a(x – h)2 + k y = a(x – h)2 + k Identify h = 3, k = 22 so vertex = (h,k) = (3,22)

Example: b) Find the ‘x’ and ‘y’ – intercepts. y – intercept can be found from the given equation ‘c’ = y –intercept so y-intercept = 4 to find x-intercepts: let y = 0 and get: 0 = -2(x – 3)2 + 22 - 22 = -2(x – 3)2 11 = (x – 3)2 + √11 = x – 3 3 + √11 = x Example: Find the equation of the quadratic function ‘’f’ with f(-1) = - 7 and a maximum value of f(2) = -1 f(-1) = - 7  (-1,-7) A Maximum value at f(2) = -1 means (2,-1) is the vertex (h,k)

so using h = 2 and k = - 1 gives us this working format: y = a(x – 2)2 + (-1)  using the other point given (-1,-7) for x and y gives us: - 7 = a(-1 – 2)2 - 1  solve for ‘a’ + 1 + 1 - 6 = a(-3)2 - 6 = 9a  divide by 9 - ⅔ = a So putting it all together  y = -⅔(x – 2)2 - 1 b) Show that the function ‘f’ has no x-intercepts. The parabola must open downward since a = - ⅔ and since the vertex is located below the x-axis at (2,-1) it cannot cross the x-axis!

Example: Where does the line y = 2x + 5 intersect the parabola y = 8 – x2 Check out the solution for example 3 found on page 40! (Show both the algebraic and graphical approach)

Example: Find an equation of the function whose graph is a parabola with x-intercepts ‘1’ and ‘4’ and a y-intercept of ‘- 8’ Look over example 4 found on page 40. Homework: page 40 CE #1-6 all; page 41 WE #1-25 left column