How do you find the vertex and intercepts for #f(x)=3 -8x -4x^2#?

1 Answer
Mar 26, 2016

#y_("intercept") = "the constant "=3#

#x_("intercpts") -> x~~0.3229" or "x~~ -2.3229#

#"Vertex "->" "(x,y)=(-1,7)#

Explanation:

Given:#" "y=-4x^2-8x+3#

This equation does not have whole numbers as roots so the formula has to be used to solve for #x_("intercepts")#.

#color(blue)("Determine the y-intercept")#

#color(green)(y_("intercept") = "the constant "=3)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex")#

Write the equation as:
#" "y=-4(x^2 +2x)+3->#part way to vertex form

Consider #2" from "2x#

Apply #(-1/2)xx2 = -1#

#color(blue)(x_("vertex")=-1)#

By substitution:

#color(green)(y_("vertex")=-4(-1)^2-8(-1)+3 = +7)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "x_("intercepts"))#

Using: #y=ax^2+bx+c# where #" "a=-4"; " b=-8"; " c=3#

#" and "x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(+8+-sqrt((-8)^2-4(-4)(3)))/(2(-4)) #

#x=(+8+-sqrt(112))/(-8) #

#x=(8+-sqrt(2^2xx28))/-8#

#x=(8+-2sqrt(28))/-8#

#color(green)(x=-1+-sqrt(28)/4 ~~0.3229" or "x~~ -2.3229)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B