calculus
posted by lana .
find the point(s) on the graph of y = (2x4 + 1)(x  5) where the tangent line slope is 1.

I assume you meant
y = (2x^4 + 1)(x  5)
using the product rule
dy/dx = (2x^4+1)(1) + 8x^3(x5)
simplify and set it equal to 1 gives me
2x^4 + 1 + 8x^4  40x^3 = 1
10x^4  40x^3 = 0
10x^3(x4) = 0
so x = 0 or x = 4
those are the x's of your two points where the slope of the tangent is 1
sub into the origianal to get the y's
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