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Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:
So we apply chain rule:
=
Set our first derivative to zero and solve for x:
3(x^2 - 1) * 2x = 0
So we can see that (by plugging in) 0, -1 and 1 makes our solution true
So our critical value is x = 0, x = -1, x = 1
So we apply chain rule:
=
Set our first derivative to zero and solve for x:
3(x^2 - 1) * 2x = 0
So we can see that (by plugging in) 0, -1 and 1 makes our solution true
So our critical value is x = 0, x = -1, x = 1
The answer is that the equation y=5x+31 is perpendicular to the equation y=-1/5x+3 and that equation passes through (-5,6) also y=5x+31 is in y=mx+b form already
y=mx+b
y=mx+b