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Let P be a point on the graph of y=e−x2/16 with NONZERO x-coordinate a. The normal line to the graph through P will have y-intercept b. (Note: The normal line at P is perpendicular to the tangent line at P.)

Verified Answer :

maxday

Yes. Just looking at the figure should confirm your conclusion: as P approaches the y-axis, the slope of the the normal becomes vertical and b is completely undetermined.

There is ambiguity in where the "3" goes in the exponent. You need more parentheses or curly brackets: 
e^{-x^2/3} or e^{-x^{2/3}}
when expressed as LaTeX code, becomes
e−x2/3 or e−x2/3

The deduction that could be made regarding the given intercepts would be:

  • Yes, the image provides the deduction that the normal slope would turn vertical and b would remain uncertain in case P moves towards axis y.
  • In this situation, the place where the exponent becomes 3, it becomes;
  • \(e^{-x^2/3}\) or \(e^{-x^{2/3}}\)
  • Thus, \(e^{-x^2/3}\) or \(e^{-x^{2/3}}\)is the correct answer.

Learn more about 'Perpendicular' here:

brainly.com/question/11707949

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