Asked by amir
x^2+8x+32 its critical points in derivatives
Answers
Answered by
oobleck
y = x^2 + 8x + 32
y' = 2x + 8
critical points are where y' = 0
But you already know what a parabola looks like, so don't forget your Algebra I now that you're learning calculus!
y' = 2x + 8
critical points are where y' = 0
But you already know what a parabola looks like, so don't forget your Algebra I now that you're learning calculus!
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