Find an equation of the tangent line to the curve at the given point.

y = SQRT(x) , (16, 4)

To find the equation of a line, we need the slope of the line and a point on the line.

Since we are requested to find the equation of the tangent line at the point (16, 4), we know that (16, 4) is a point on the line. So we just need to find its slope.

The slope of a tangent line to f(x) at x = a can be found using the formula

mtan = lim x→a f(x)-f(a)/ x-a

In this situation, the function is f(x) =

and a =

Consider the parabola y = 6x − x2.

(a) Find the slope of the tangent line to the parabola at the point (1, 5).

(b) Find an equation of the tangent line in part (a).
y =

If a ball is thrown into the air with a velocity of 50 ft/s, its height (in feet) after t seconds is given by y = 50t− 16t2. Find the velocity when t = 2.

Let

h = s(t) = 50t − 16t2

give the height of the ball at time t.

Then the ball's velocity at time t = a can be found by

v(a) = lim t→a s(t)-s(a) / t-a

We are requested to find the velocity at t = 2; therefore we use a = and have

v(2)= Lim t→2 s(t)-s(2) / t-2

lim t→2 (50t − 16t^2) - (50()-16(2)^2) / t-2

lim t→2 (50t − 16t^2) - () / t-2

Find f '(a).

f(x) = 4x2 − 5x + 4

f'(a)=

(a) Find the slope m of the tangent to the curve y = 7 + 4x2 − 2x3 at the point where x = a.
m =

(b) Find equations of the tangent lines at the points (1, 9) and (2, 7).

y(x) = (at the point (1, 9))
y(x) = (at the point (2, 7))

(c) Graph the curve and both tangents on a common screen.

Find f '(a).

f(t) = t4 − 5t

f '(a) =