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given y=4x^2-2x+(2), find the equation of the tangent line that is horizontal to the curve
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You can differentiate it first
and you get 8x-2
what do you do after
you can try equating dy/dx=0 then solve for x
dy/dx=8x-2 8x-2=0 x=1/4
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so y=4x^2-2x+(2) y=4(1/4)^2-2(1/4)+2 y=7/4
how do we put it in slope intercept form?
then use \[y-y_1=m(x-x_1)\]
whats m=?
so the final answer is?
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equation of tangent line is y=7/4
thanks!
can you help me with one more problem?
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