Question

Find the slope of the graph of the function y=5x/x-2 at (3,15). Then find an equation for the line tangent to the graph at that point.

Answer 1

The slope of the graph at the point (3,15) is -10. The equation of the tangent line at that point is y = -10x + 45.

Step-by-step explanation:

To find the slope of the graph, we need to differentiate the equation y=5x/(x-2). Using the quotient rule, the derivative of this equation is (5(x-2) - 5x)/(x-2)². Simplifying this further, we get -10/(x-2)². Now, substitute the x-coordinate of the point (3,15) into the derivative to find the slope at that point. Substituting x=3, we get a slope of -10/1 = -10.

To find the equation of the tangent line at the point (3,15), we use the point-slope form of a line, y - y1 = m(x - x1). Substituting the slope (-10) and the coordinates of the point (3,15) into this equation, we get y - 15 = -10(x - 3). Simplifying further, we get y - 15 = -10x + 30. Rearranging the equation, we get y = -10x + 45. Therefore, the equation of the tangent line to the graph at the point (3,15) is y = -10x + 45.

Answer 2

The first derivative gives you the formula for the slope of the tangent line to the curve at a given point.

Find the first derivative of the function y=5x/x-2:

(x-2)(5) - (5x)(1)
dy/dx = ------------------------
(x-2)^2

simplify the algebra here.

(3-2)(5) - (15)(1)
Then, at the point (3,15), dy/dx = -------------------------- = -10
(3-2)^2

What is the equation of the tangent line? In other words, what is the eqn of the tan line to the given curve at when x = 3, if the slope is -10?

y-15 = -10(x-3)

This can be re-written in other forms if desired.