For #f(x) =4-5x^2#, what is the equation of the line tangent to #x =2/5#?

1 Answer
Bill K.
Nov 9, 2015

The equation of the line tangent to the graph of #f(x)=4-5x^2# at #x=2/5# is #y=16/5-4(x-2/5)=-4x+24/5#.

Explanation:

The derivative is #f'(x)=-10x# and the slope of the tangent line is therefore #f'(2/5)=-10 * 2/5 = -4#.

Since #f(2/5) = 4-5 * 4/25=4-4/5=16/5#, it follows that the equation of the line tangent to the graph of #f(x)=4-5x^2# at #x=2/5# is #y=16/5-4(x-2/5)=-4x+24/5#.

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