How do you find the important points to graph #y = (x - 3)² + 1#?

1 Answer
Nov 2, 2015

Answer:

See explanation

Explanation:

Standard form of the equation is
#y=ax^2+bx+c#
Note that #a# could be of value 1
If #a# is positive then the graph is a horse shoe shape with the curve at the bottom. If negative then the other way round.

Square the bracket and then add the 1 giving:
#y=x^2 - 6x +4#
Tony B

Using standard form equation

#x" of "(x,y)_("minimum")" is "(-1/2) times b/a#

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So for your case we have:

#color(green)(x" of "(x,y)_("minimum") =(-1/2) times (-6) = color(green)(+3))#

#color(red)("substitute "x=3" in your equation to find "y_("minimum"))#

#y = (3)^2 -6(3)+4#
#y =13 - 18#
#y = -5#
#color(green)(y" of " (x,y)_("minimum")=(-5)#

Putting it all together

#color(green)( (x,y)_("minimum")=(3,-5)#
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#color(green)("To find x-intercepts substitute y=0 and solve")#

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#color(green)("To find y-intercepts substitute x=0 and solve")#

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