#color(green)("Method tutorial")#
It is a quadratic curve (Horse shoe shape)
#color(blue)("General shape")#
The coefficient of #x^2# is +1 #-> positive#
so the shape is U. If the coefficient had been negative it would have been the other way up ( #nn#)
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#color(blue)("The y-intercept")# is when #x=0# so you can find it by substitution. (Tip: it is the constant!!!!)
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#color(blue)("The x-intercept")# is when #y=0# so you can find this by factoring (much the quicker method) or using the formula:
Standard form equation of #ax^2+bx+c=0#
In conjunction with # x= (-b+-sqrt(b^2-4ac))/(2a)#
If the #sqrt(b^2-4ac)# bit is negative then the curve does not cross the x-axis and if you are determined to get some answers you have to use a branch of maths called Complex Numbers.
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#color(blue)("Turning point of the curve (vertex)")#
The Top/Bottom of the curve (maximum/minimum) is quite easy to find. Using the standard form equation. Change it to
#y=a(x^2+b/ax) +c#
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#color(brown)("There another one you can use called vertex equation which has a"#
#color(brown)("different approach")#
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In your case a=1 so #b/a x ->color(blue)((-1)/1) x #
#color(green)(v_("vertex"))=(-1/2)xxcolor(blue)(b/a) -> (-1/2)xx(color(blue)(-1)) =color(green)(+1/2)#
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#color(green)(y_("vertex")) ->#substitute the found value of #color(green)(v_("vertex"))#
