Question 62769
f(x)=x^2 + x- 6

vertex and intercepts of the quadratic function 

Given f(x) = x^2 + x- 6 --->(1)

It is a quadratic equation, where a = 1, b =1 and c = -6

vertex = - b /2a = -1 /2*1 = - 1/2

x = -1/2

Now substitute for x = -1/2,

Put this value in equation(1), we get

f(-1/2) = (1-/2)^2 + (1/2) - 6

        = 1/4 + 1/2 - 6

        = 3 /4 - 6 

        = -21 /4

Therefore the vertex is given by,

(x, f(x) = ( -1/2,  -21/4)


To find the intercepts, we put x =0 in  equation(1) we get

y = f(0) =  -6

THis is the y-intercepts

Now put f(x) = 0 in the equation(1), we get
x^2 +x - 6 = 0

x^2 +3x -2x -6 = 0

x(x+3) -2(x+3) = 0

(X-2)(x+3)= 0

Therefore x = 2 or -3

The x intercepts are x =2 and x =-3