Question

Find an equation of the curve whose tangent line has a slope of

f′(x)=4x−8/9

given that the point

(−1,−7)

is on the curve.

Answer #1

The point on the curve is

That is

Equation of the curve is

------------------------------------------------------------------

Being f (x) = x^2 - 4x - 3, determine the slope of the line
tangent to the curve of
f (x) at the point where x =3.
VIII. Being f (x) = 5x^2 + 3x - 9, determine the slope of the
line tangent to the curve of
f (x) at the point where x = -1.
IX. Determine the equation of the line to the curve of f (x) =
x^2 - 9x , at the point where...

find the equation of a tangent line(s) to the curve with slope 5
f(x)=x^3 + 3x^2 - 4x - 12
f'(x)= 3x^2 +6x - 4

Find the slope of the line tangent to the curve y=x^2 at the
point (-0.9,0.81) and then find the corresponding equation of the
tangent line.
Find the slope of the line tangent to the curve y=x^2 at the
point (6/7, 36,49) and then find the corresponding equation to the
tangent line.
answer must be simplified fraction

Q1. (a) Find the equation of the tangent to the curve
f(x)=x^2+4x+2 at the point where the gradient is 8.
(b) Find the equation of the normal at this point, where normal
gradient times tangent gradient is -1.

find the slope then find the equation of a tangent line that is
tangent to the curve f(x)=x2tan(x) at (pi, 0) (hint: use
product rule)

Find an equation of the tangent line to the curve at the given
point. A) y = 6x + 3 cos x, P = (0, 3) B)y = 8 x cos x P = \(pi ,
-8 pi)
B)Find an equation of the tangent line to the curve at the given
point.
y = 8 x cos x
C)
If H(θ) = θ cos θ, find H'(θ) and H''(θ).
find H'(
θ)
and H"(θ)

. Find the slope of the tangent line to f-1 at the
point P(-1, 0) if f(x) = x+1/ x-1, and then find the
slope-intercept equation of the tangent line to the graph of
f-1 at P.

let f(x) = sqrt x^4+4x+4.find the equation of the tangent line
to the graph of f −1(a) when a = 3

Find the equation of the tangent line to the graph of the
function f(x)=(x^2+8)(x−2) at the point (1,−9).
I thought it was re-writen as (2x^2 + 8)(x-2) then plugging in 1
for x and solving. I came up withit in slope form y = -20x - 1 but
says im wrong. What steps did i miss?

Find an equation of the tangent line to the curve at the given
point.
1.) y= sqrt(5x+ 9), at x= 10.
2.) y= cos(x) + cos^3(x), at x=π/6.

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