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Find an equation of the tangent line to the graph of "f" at the given point. f(x)=x²+3, (1,4)
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do you know what the derivative of \[f(x)=x^2+3\] is?
No
do you know the power rule or the constant rule?
Yes, I do.
Then use them to find the derivative.
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Power Rule \[(x^n)'=nx^{n-1} \] Constant Rule (C is a constant) \[(C)'=0\]
\[f'(x)=(x^2)'+(3)'=?\]
2x?
yep
f'(1) will be the slope remember f'(x)=2x so what is f'(1)=?
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2
so we know this about the tangent line y=2x+b so we need one more thing, the y-intercept
You know a point on this line (1,4) Plug it in to find b, the y-intercept.
\[4=2(1)+b\] I replaced x with 1 and y with 4.
find b
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b=2
so the tangent line is y=mx+b where m=2 and b=2 y=2x+2
alright, thank you
A plot is attached.
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