I am going to use a method that is the beginning of building the Vertex Equation Form but not taking it all the way.
First observation is that #4x^2# is positive. This means the graph is of shape type #uu#
(If it had been #-4x^2# then the shape type would be #nn#)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "y=4x^2-6x-3#...........................(1)
#color(blue)("To determine "x_("vertex"))#
Write as:#" "y=4(x^2-6/4 x) -3#
#color(brown)("I have taken the 4 from "4x" outside the brackets.")#
#color(brown)("Note that "4xx(-6/4) x=-6x)#
Now consider the #-6/4" from "-6/4x#
Apply this operation: #(-1/2)xx(-6/4) = +3/4#
#" "color(blue)(x_("vertex") = +3/4#
#color(brown)("With some equations you can do this in your head making the")##color(brown)("process very fast!")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine "y_("vertex"))#
Substitute#" "color(green)(x=3/4)# into equation (1)
#color(brown)(y=4x^2-6x-3" "->" "y=4(color(green)(3/4))^2-6(color(green)(3/4))-3)#
#color(blue)(" "y_("vertex")=9/4-9/2-3 =- 5 1/4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine "y_("intercept"))#
#color(brown)("The y intercept is when "x=0)#
So equation (1) becomes
#y=4(0)^2-6(x)-3#
#color(blue)(y_("intercept")=-3)#
#color(magenta)("Vertex "->" "(x,y)" "->" "(3/4,-21/4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine "x_("intercepts"))#
The values for x are non integer values so use the formular
Standard for #y=ax^2+bx+c#
where #x =(-b+-sqrt(b^2-4ac))/(2a)#
#a=+4#
#b=-6#
#c=-3#
Giving:
#" "x=(+6+-sqrt((-6)^2-4(4)(-3)))/(2(4)) #
#" "x=(+6+-sqrt(36+48))/8 #
#" "x=(6+-sqrt(2^2xx21))/8#
#" "x=(6+-2sqrt(21))/8#
#color(blue)(" "x_("intercepts")~= 1.896" or " -0.396" to 3 decimal places")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
