How do you find the important points to graph #F(x)= 2x^2 - 4x +1#?

1 Answer
Nov 1, 2015

See explanatioin

Explanation:

Tony B

#color(green)("Standard form for a quadratic is: "ax^2 + bx +c)#

#color(green)("Convert this into " a(x^2 +b/a+c/a))#

#color(green)("In your case we factorise it into")#

#color(green)(y= 2(x^2 -2x+1/2) )#

#color(green)("It is important that there is no coefficient directly in front of the "x )#

In this case we look at what is inside the brackets to get what we want.

#x^2# is positive: that gives you an upwards horse shoe shape.
#" "# If it had been negative then the horse shoe would
#" "# have been the other way up.
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#color(blue)("The maxima/minima x value is " (-1/2) times b/c)#

So for you it will be at #color(red)( x_("minima")) =(-1/2) times (-2 )=color(red)(1)#
To find the y value at the minima substitute #(-1/2) times (-2)=(1)# for x

So
#y =2x^2 - 4x + 1#
becomes:
#color(red)(y_("minima")) =2(1)^2-4(1)+1 = color(red)( -1)#

Thus:

# color(red)((x,y)_("minima"))# #color(blue)( -> (1,-1))#
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#color(blue)("To find y intercept substitute " x = 0" and solve")#

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#color(blue)("To find x intercept substitute "y=0" and solve")#

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