yes
take the derivative and sub x=1 into that, giving you the slope of the tangent
also sub x=1 into the original equation to get y, so then you have the slope of the tangent and the point of contact.
Can you find the equation of a line given the slope and a point on it??
BTW, is that supposed to say.. 5x^2 +4 ?
find the equation of a line tangent to the graph of y = x / (square root of 5x^s +4), at x = 1.
am i suppose to take the derivative? if so, then what else?
1 answer
1 answer