Suppose that fx x2 4x 12 a What is the vertex of f b What

Suppose that f(x) = x^2 - 4x - 12 a. What is the vertex of f? b. What arc the x-intercepts of the graph of f? Find a quadratic function in the form f(x) = ax^2 + bx + c:whose vertex is (-1, 5) and y-intercept is 3. Solve the following inequalities:. Express the solution in interval form. a. x^2 - 30 greaterthanorequalto 13x b. x + 3/x-4 greaterthanorequalto 0

Solution

1) f(x) = x^2 -4x -12

Vertex : ax^2 +bx +c given by x = -b/2a

x = 4/(2*1) = 2

f(2) = 4 -4*2 -12 = -16

Vertex ( 2, -16)

X intercepts: x^2 - 4x - 12 =0

x^2 - 6x +2x -12 =0

x( x-6) +2(x -6) =0

(x+2)(x-6) =0 = -----> x= -2 , 6

X intercepts : ( -2, 0) and ( 6, 0)

2) ax^2 +bx +c , vertex ( -1 , 5) and y intercept is 3

so, y = ax^2 +bx +c

x=0 ; y =c = 3

c =3

So, y = ax^2 +bx + 3

vertex : x = -b/2a

-1 = -b/2a ----> b = 2a

(-1 , 5) lies on ax^2 +bx +c =y

5 = a -b

5 = a -2a ----> a = -5

b = -10

f(x) =-5x^2 -10x +3