SOLUTION: Find the slope of the line tangent to the following at point, where x=2 f(x)=e^(-0.1x^2+2x-3).

Algebra ->  Graphs -> SOLUTION: Find the slope of the line tangent to the following at point, where x=2 f(x)=e^(-0.1x^2+2x-3).       Log On


   

Question 1155282: Find the slope of the line tangent to the following at point, where x=2
f(x)=e^(-0.1x^2+2x-3).
Found 2 solutions by MathLover1, Alan3354:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Find the slope of the line tangent to the following at point, where x=2
f%28x%29=e%5E%28-0.1x%5E2%2B2x-3%29.......if x=2
f%28x%29=e%5E%28-0.1%2A2%5E2%2B2%2A2-3%29
f%28x%29=e%5E%28-0.1%2A4%2B4-3%29
f%28x%29=e%5E%28-0.4%2B1%29
f%28x%29=e%5E%280.6%29
so, the point is (2,e%5E0.6) or (2,1.8)

compute slope m
take first derivative
df%28x%29%2Fdx=e%5E%28-0.1x%5E2%2B2x-3%29%2A%28-0.2x%2B2%29
if x=2
m=e%5E%28-0.1%2A2%5E2%2B2%2A2-3%29%2A%28-0.2%2A2%2B2%29
m=2.91539

y=mx%2Bb
y=2.91539x%2Bb........use point (2,e%5E0.6)to find b
e%5E0.6=2.91539%2A2%2Bb
b=e%5E0.6-5.83078
b=1.8221188003905089-5.83078
b=-+4.00866
and, tangent is:
y+=+2.91539x+-+4.00866



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the slope of the line tangent to the following at point, where x=2
f(x)=e^(-0.1x^2+2x-3).
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f(x)=e^(-0.1x^2+2x-3)
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f'(x) = e%5E%28-0.1x%5E2%2B2x-3%29%2A%28-0.2x+%2B+2%29
Sub 2 for x
===================
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f'(2) = e%5E%28-0.4+%2B+4+-3%29%2A%28-0.8+%2B+2%29
f'(2) = 1.2e%5E%280.6%29