Given: #y=x^2-4x+2" ".....................Equation(1)#
#color(blue)("They may not have shown you this trick.")#
Consider the standardised form #y=ax^2+bx+c#
Write this as #y=a(x^2+b/ax)+c #
This is part way to completing the square.
Then #x_("vertex")=(-1/2)xxb/a# which in this case is:
#x_("vertex")=(-1/2)xx(-4)/1 =+2#
Then by substitution you can dtermine #y_("vertex")#
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#color(blue)("How they may expect you to do it by completing the square")#
#color(brown)("Determine the vertex")#
Given that #y=ax^2+bx+c" then "y=a(x+b/(2a))^2+c#
But this has introduces a value that is not in the original equation so include a correction. I choose #k# giving:
#y=a(x+b/(2a))^2+k+c#
Set #a(b/(2a))^2+k=0# neutralising the error.
For this question we have:
#color(green)(y=1(x+ (-4)/(2xx1))^2+color(red)(k)+2)" ".................Equation(2)#
#color(green)("Set "((-4)/2)^2+color(red)(k)=0 => color(red)(k)= -4)#
#color(green)(y=(x-2)^2color(red)(-4)+2)#
#color(green)(y=(xcolor(black)(-2))^2color(darkviolet)(-2))" ".......Equation(3)#
#color(green)(x_("vertex")=(-1)xx(color(black)(-2))=+2)#
#color(green)(y_("vertex")=color(darkviolet)(-2))#
#"Vertex "->(x,y)=(2,-2)#
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#color(brown)("Determine the y -intercept")#
Consider the original equation: #y=x^2-4xcolor(red)(+2)#
#y_("intercept") = color(red)(+2)#
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#color(brown)("Determine the x -intercepts")#
The #x^2# term is positive so the graph is of form #uu#
#y_("vertex") = -2# so the graph crosses the x-axis so there are two solution for #y=0=(x-2)^2-2#
Add 2 to both sides
#2=(x-2)^2#
Square root both sides
#+-sqrt2=x-2#
#x=2+-sqrt(2)#
#x~~3.414213.... -> x~~3.41# to 2 decimal places
#x~~0.585786.... -> x~~0.59 # to 2 decimal places
