SOLUTION: Consider the following function f(x)=x^2-sqrt x find the equation of the line tangent to the graph at the point (9,78) y(x)= can someone help please

Question 200393: Consider the following function
f(x)=x^2-sqrt x
find the equation of the line tangent to the graph at the point (9,78)
y(x)=
can someone help please
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Consider the following function
f(x)=x^2-sqrt x
find the equation of the line tangent to the graph at the point (9,78)
----------------
The slope m is 2x - (1/2)x^(-1/2)
m = 2x - (1/2sqrt(x))
At x = 9, m = 2*9 - (1/6)
m = 107/6
------------
y-78 = (107/6)*(x-9)
y = 107x/6 - 107*3/2 + 78
y = 107x/6 - 165/2 slope-intercept from
---------------------
6y - 468 = 107x - 5778
107x - 6y = 5310 standard form

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