How do you identify the important parts of #y= 2x^2+7x-21# to graph it?
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"How does implicit differentiation work?"
y intercept = -21 ; shape is an upward horse shoe with the minimum at #(1 3/4, - 1 3/4)# Crosses the x axis at y=0 so factorise and equate to 0
equation standard form is #y =ax^2 + bx +c#
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in this case:
#+x^2# is upwards horse shoe shape
#-x^2# is downward horse shoe shape
So this an upwards with a minimum y value but goes on increasing from there to infinity.
y intercept at #x=0# so by substitution #y=-21#
which is the value of the constant c.
The value of #x# at the minimum may be found by completing the square ( changing the way the equation looks without changing its intrinsic value). We need to end up with both the x^2 and the x part together inside the brackets. In this case we have:
write #y=2x^2 + 7x -21# as:
#y = 2(x^2 + 7/2)^2 - 21 -("correction bit of" (7/2)^2)#
the correction is to compensate for the constant we have introduced. We still have the -21 but we have introduced #(7/2)^2# due to #7/2 times 7/2# from #( x^2 + 7/2)(x^2+7/2)#
Now we look inside the brackets at the #7/2x# part
The minimum occurs at #x_(min) = -1/2 times 7/2 = - 1 3/4#
To find #y_(min)# we substitute the found value for #x_(min)# in the original equation.
ABR again could not correct my answer. The value of #x_(min)# is 1/ 3/4# Sorry about that!!!
