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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