Let f (x) = x^2 - 6x - 27

enter the x - intercepts of the quadratic function in the boxes.

x = and x =

x = 9

x = 3

Step-by-step explanation:

find the roots of x² - 6x - 27 = 0

by factorization (or use any of your favorite methods for solving quadratic equations):

x² - 6x - 27 = 0

(x-9)(x+3) = 0

Hence,

(x-9) = 0  > x = 9

or

(x+3) = 0  > x = -3

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

y=mx+b is the final answer!   : )