Given: " "x^2-2x+1=18
Subtract 18 from both sides giving:
" "x^2-2x-17=0.........................(1)
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This one is more straight forward as the coefficient of x^2 is 1
color(blue)("Step 1")
Write as:
(x^2-2x)-17=0
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 2")
color(brown)("At this point we will change the intrinsic value of the equation so it") color(brown)("will need to have a correction added later.")
Take the power to be outside the brackets
(x-2x)^(color(red)(2))-17
Remove the x from the -2x leaving just the color(red)(-2)
(xcolor(red)(-2))^2-17
halve the -2 that is inside the brackets so that you have color(red)(-1)
(xcolor(red)(-1))^2-17
color(brown)("Now we add the correction")
color(green)("This takes the equation back to its original value so we can once again equate it to zero")
Let color(red)(k) be some constant
(x-1)^2color(red)(+k)-17=0 ................................(2)
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color(blue)("Step 3")
color(brown)("By expanding the brackets and comparing to the original equation")color(brown)("we can determine the appropriate value of "k)
Expanding equation (2)
cancel(x^2)-cancel(2x)+1+k-cancel(17) " "=" "cancel(x^2)-cancel(2x)-cancel(17)
k=-1 so equation (2) becomes:
color(blue)((x-1)^2-18=0) .................................(3)
By comparing the values in the vertex format equation you can see how to obtain the vertex coordinates.
Multiply the constant inside the bracket by (-1) to get x_("vertex") giving 1 and y_("vertex") can be read directly as -18
color(blue)("vertex "-> (x,y)" "->" "(1,-18))
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color(blue)("Step 4")
y-intercept at x=0
=>(x-1)^2-18=y" "->" "(0-1)^2-18=y
color(blue)(y_("intercept")= -17)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Step 5")
x-intercepts when y=0 so from equation (3)
sqrt( (x-1)^2)=sqrt(18)
x-1=+-3sqrt(2)
x=1+-3sqrt(2)
color(blue)(x_("intercept")~= + 5.24" "or" "-3.24 to 2 decimal places
