Question

Please verify my answer. Do this look correct to you? If not please explain

Answer 1

Answer:

  y = e·x

Step-by-step explanation:

The equation of a line tangent to a curve at a point is conveniently written in the point-slope form. The slope is the derivative of the function at the point. For your function, the rules applicable to products and exponential functions apply.

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  y = (e^x)(x^2 -2x +2

  y' = (e^x)(x^2 -2x +2) +(e^x)(2x -2) . . . . . (uv)' = u'v +uv', (e^x)' = e^x

  y' = (e^x)x^2 . . . . . simplify

For x=1, the slope is ...

  y' = (e^1)(1^2) = e

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The point-slope form of the equation for a line is ...

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

  y -e = e(x -1) . . . . . . . line with slope 'e' through point (1, e)

  y = e·x -e +e . . . . . add e

  y = e·x . . . . . . . . . . equation of the tangent line

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Additional comment

It is often the case that ex is written when e^x is intended. We are trying to avoid that ambiguity here by writing the equation with an explicit "times" symbol.