Find the point on the graph of y=x^2 where the tangent line has slope m= - 1/3

Here the graph y=x²

Taking the 1st derivative here of the y it will be

y1 = 2x

m=-1/3

That is m=2x

so x=-1/6

so y=(-1/6)²=1/36

So the coordinate of point will be=(-1/6 , 1/36)

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!