#color(blue)("Formula type 1")#
When I was in my upper years of school I made a point of writing down the quadratic formula each time I needed to use it. The repetition of this act really fixed it into my memory. There was a lot of repeats and it is the repetition of #ul("writing it down")# that did the trick. I am a lot older and I can still remember it.
Given the standard form of #y=ax^2+bx+c# then if you set #y=0# we have for that condition: #x=(-b+-sqrt(b^2-4ac))/(2a)#
As long as the part #b^2-4ac# is positive and greater than 0 the graph crosses the x-axis ( 2 values for #x#).
If the part #b^2-4ac = 0 # then the x-axis forms a tangent to the vertex.
If #b^2-4ac# is negative then the graph does not cross the x-axis and you are entering the realm of what is called complex numbers.
If the part #ax^2# is positive then the graph is of general shape #uu#
If the part #ax^2# is negative then the graph is of general shape #nn#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Formula type 2")#
There is another format which has the same values as the previous one (on the whole) and again you need to repeatedly write this one down. It is the Vertex equation.
Given #y=ax^2+bx+c" then "y=a(x+b/(2a))^2+k_1+c#
We need #k_1# as changing into this format introduces a value that is not in the type #y=ax^2+bx+c#. So we remove this by setting
#k_1+a(b/2a)^2 =0# We then add this to #c# giving #k_2#:
#y=a(x+b/(2a))^2+k_2" where "k_2=k_1+c#
#x_("vertex")->(-1)xx(b/(2a)) = -b/(2a)#
#y_("vertex")=k_2#
Determine the x-intercepts you set #y=0=a(x+b/(2a))^2+k_2# and solve for #x#.
