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#)

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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!")#

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#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)#

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#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)#

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#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")#

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