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

Question

Ethan Adkins · Accepted Answer

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

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

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

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)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Eva Abramson · Answer

undefined To find the vertex of a quadratic function in the form $ f(x) = ax^2 + bx + c $, use the formula $ x = -\frac{b}{2a} $. 
For the given function $ f(x) = 3 - 8x - 4x^2 $, 
$ a = -4 $ and $ b = -8 $.
So, $ x = -\frac{-8}{2*(-4)} = -\frac{-8}{-8} = 1 $.
To find the y-coordinate of the vertex, substitute $ x = 1 $ into the function: $ f(1) = 3 - 8(1) - 4(1)^2 = 3 - 8 - 4 = -9 $.
Therefore, the vertex is $ (1, -9) $.
To find the x-intercepts, set $ f(x) = 0 $ and solve for $ x $.
$ 3 - 8x - 4x^2 = 0 $.
Using factoring or the quadratic formula, you can find the roots or x-intercepts.