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1 answer:
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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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<em>Additional comment</em>
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.
