Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical points for #y= 2x^2 + 10x - 7#?

Answer 1

#x=-5/2#

The critical points of a function are the numbers that make its first derivative equal to zero #=># for #f(x)#, find #f'(x)# and equate it to #0#. The points that make #f'(x)=0# are called "critical points. "

#y=2x^2+10x-7#

#y'=4x+10#

#y'=0=>4x+10=0#

#=>4x=-10#

#=>2x=-5#

#=>x=-5/2# is the critical point.

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Answer 2

Paisley Johnson

To find the critical points for ( y = 2x^2 + 10x - 7 ), you need to first find the derivative of the function, ( y' ), and then solve for ( x ) where ( y' = 0 ).

First, take the derivative of ( y ) with respect to ( x ) (denoted as ( y' )): [ y' = \frac{dy}{dx} = 4x + 10 ]

Now, set ( y' ) equal to ( 0 ) and solve for ( x ): [ 4x + 10 = 0 ] [ 4x = -10 ] [ x = -\frac{10}{4} = -\frac{5}{2} ]

So, the critical point is ( x = -\frac{5}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.