Math NS FUNCTIONS QUADRATIC

y = f (x) = ax 2 + bx + c Any function of the form where a  0 is called a Quadratic Function

a b c f(x) = 3x 2 – 2x – 5 = 3, = -5 = -2, Example: Consider the quadratic function in general form: f(x) = 3x 2 – 2x – 5 a b c = 3, = -2, = -5 Note that if a = 0 we simply have the linear function f(x) = bx + c

Consider the simplest quadratic equation y = x 2 Here a = 1, b = 0, c = 0 Plotting some ordered pairs (x, y) we have: y = f (x ) = x 2 x f (x ) (x, y ) -3 9 (-3, 9) -2 4 (-2, 4) -1 1 (-1, 1) 0 0 (0, 0) 1 1 (1, 1) 2 4 (2, 4) 3 9 (3, 9)

y = x2 (x, y) (-3, 9) (-2, 4) (-1, 1) (0, 0) (1, 1) (2, 4) (3, 9) -3 -2 -1 1 2 3 Vertex (0, 0) A parabola with the y-axis as the axis of symmetry.

Graphs of y = ax 2 will have similar form and the value of the coefficient ‘a ’ determines the graph’s shape. y y = 2x 2 y = x 2 4 3 2 1 y = 1/2 x 2 a > 0 opening up -3 -2 -1 1 2 3 x

a < 0 opening down y = -2x 2 In general the quadratic term ax 2 in the quadratic function f (x ) = ax 2 +bx + c determines the way the graph opens.

Consider f (x ) = ax 2 +bx + c In a general sense the linear term bx acts to shift the plot of f (x ) from side to side and the constant term c (=cx 0) acts to shift the plot up or down. y a > 0 x-intercept c Notice that c is the y -intercept where x = 0 and f (0) = c a < 0 x c y-intercept Note also that the x -intercepts (if they exist) are obtained by solving: y = ax 2 +bx + c = 0

(1) Opening up (a > 0), down (a < 0) It turns out that the details of a quadratic function can be found by considering its coefficients a, b and c as follows: (1) Opening up (a > 0), down (a < 0) (2) y –intercept: c (3) x -intercepts from solution of y = ax 2 + bx + c = 0 You solve by factoring or the quadratic formula (4) vertex =

Plug in x = to find the y-value Example: y = f (x ) = x 2 - x - 2 here a = 1, b = -1 and c = -2 (1) opens upwards since a > 0 (2) y –intercept: -2 (3) x -intercepts from x 2 - x - 2 = 0 or (x -2)(x +1) = 0 x = 2 or x = -1 Plug in x = to find the y-value (4) vertex:

y (-1, 0) (2, 0) x -2 -1 0 1 2 y = x 2 - x - 2 -1 -2 -3

Example: y = j (x ) = x 2 - 9 here a = 1, b = 0 and c = -9 (1) opens upwards since a > 0 (2) y –intercept: -9 (3) x -intercepts from x 2 - 9 = 0 or x 2 = 9  x = 3 (4) vertex at (0, -9)

y (-3, 0) (3, 0) x -3 0 3 y = x 2 - 9 -9 (0, -9)

Example: y = g (x ) = x 2 - 6x + 9 here a = 1, b = -6 and c = 9 (1) opens upwards since a > 0 (2) y –intercept: 9 (3) x -intercepts from x 2 - 6x + 9 = 0 or (x - 3)(x - 3) = 0  x = 3 only (4) vertex:

y (0, 9) 9 y = x 2 - 6x + 9 (3, 0) x 3

Example: y = f (x ) = -3x 2 + 6x - 4 here a = -3, b = 6 and c = -4 (1) opens downwards since a < 0 (2) y –intercept: -4 (3) x -intercepts from -3x 2 + 6x - 4 = 0 (there are no x -intercepts here) (4) vertex at (1, -1) Vertex is below x-axis, and parabola opens down!

y x 1 2 (1, -1) -1 -4 y = -3x 2 + 6x - 4 (0, -4)

ax 2 + bx + c = 0 The Quadratic Formula It is not always easy to find x -intercepts by factoring ax 2 + bx + c when solving ax 2 + bx + c = 0 Quadratic equations of this form can be solved for x using the formula:

Example: Solve x 2 − 6x + 9 = 0 here a = 1, b = -6 and c = 9 Note: the expression inside the radical is called the “discriminant” Note: discriminant = 0 one solution as found previously

Example: Solve x 2 - x - 2 = 0 here a = 1, b = -1 and c = -2 Note: discriminant > 0 two solutions

Example: Find x -intercepts of y = x 2 - 9 Solve x 2 - 9 = 0 a = 1, b = 0, c = -9 Note: discriminant > 0 two solutions x = 3 or x = -3

Example: Find the x -intercepts of y = f (x) = -3x 2 + 6x - 4 a = -3, b = 6 and c = -4 Solve -3x 2 + 6x - 4 = 0 Note: discriminant < 0 no Real solutions  there are no x -intercepts as we discovered in an earlier plot of y = -3x 2 + 6x - 4

FUNCTIONS QUADRATIC The end.