Question

Let f(x) =\sqrt{26-x} .. The slope of the tangent line to the graph of f(x) at the point (1,5) is..??

Answer 1

(-1/2)(26-x)^*-1/2) is the equation for slope. Plug in x = 1

Answer 2

(-1/2)/sqrt(25) = -1/10

Answer 3

as long as u meant f(x) = sqrt(26-x)

Answer 4

yes bahrom7893, that is what I met..

Answer 5

ok, the question continues:
The equation of the tangent line to the graph of f(x) at (1,5) is y=mx+b for
m=
and
b=

Answer 6

y-5=M(x-1)
y-5=(-1/10)(x-1)
y = (-1/10)x + (51/10)

Answer 7

ok.. I am soo lost. Let me post the whole question, hold on..

Answer 8

Let f(x) =\sqrt{26-x}
The slope of the tangent line to the graph of f(x) at the point (1,5) is .......????
The equation of the tangent line to the graph of f(x) at (1,5) is y=mx+b for
m=......????
and
b=......??????
Hint: the slope is given by the derivative at x=1

Answer 9

m = (-1/10)
b = (51/10)

Answer 10

Thanks!! what about this one:
Let f(x) =25/x The slope of the tangent line to the graph of f(x) at the point (-6,-25/6) is __________???? The equation of the tangent line to the graph of f(x) at (-6,-25/6) is y=mx+b for m=__________????
and
b=_____????