Tangent line and their slopes?

Question

Find an equation for the line tanget to the curve at the given point. Then sketch the curve and tangent line together.

y=9*sqrt(x) , (9,27)

Answer 1

The curve y = 9√x

Apply derivative on each side with respect of x.

y' = 9 (1/2√x)

Substitute x = 9

y' = 9(1/2√9)

y' = 9(1/6)

y' = 3/2

This is the slope of tangent line to the curve at (9, 27).

To find the tangent line equation, substitute the values of m = 3/2 and (x, y ) = (9, 27).  in the slope intercept form of an equation y = mx + b

27 = (3/2)9 + b

27 = 27/2 + b

b = ( 54 - 27)/2

b = 27/2

Tangent line equation is y = (3/2)x + 27/2.

Graphs of curve and line

• Liner function y = (3/2)x + 27/2

y-intercept is 27/2, so the line crosses the y-axis at (0, 27/2)

Using slope find the next point.

Slope = rise/run = 3/2

Start at point (0, 27/2), move 3 units down and 2 units right, then plot the point (2,16.5).

Draw a line through these points.

• Radical function y = 9√x

Domain of the function y = 9√x is all positive numbers.

Choose values for y and find the corresponding values for x (positive).

y	y = 9 √x	(x, y)
0	y = 9 √0	(0, 0)
3	y = 9√3	(3, 15.58)
6	y = 9√6	(6, 22.04)
9	y = 9√9	(9, 27)

Draw the curved line through the above points, remembering not to extend the line to the left of x  = 0.

Graph