The line tangent to the graph of function f at the point (8,1) intersects the y-axis at y=3. How do you find f'(8)?

Question

The line tangent to the graph of function f at the point (8,1) intersects the y-axis at y=3.  How do you find f'(8)?

Isaiah Allen · Accepted Answer

Write the equation of the line tangent to the graph at #(8,1)#. Substitute for #x# and #y#, then solve for #f'(8)#

You have choices for how to write the equation of the tangent line.

Point Slope Form solution

The tangent line contains point #(8,1)# and has slope #f'(8)#, so its equation is

#y-1=f'(8)(x-8)#

The line contains #(0,3)#, so we get

#3-1=f'(8)(0-8)# which leads to

#f'(8) = -1/4#

Slope-Intercept Form solution

The tangent line has slope #f'(8)# and #y#-intercept #3#, so the equation of the tangent line is

#y=f'(8)x+3#

We know that the point #(8,1)# is on the line, so we get

#1=f'(8)*(8)+3#.

This leads to #f'(8) = -1/4#

Charles Anctil · Answer

#f'(8)# is the slope of the tangent line at #x=8#.

We know that the tangent line at #x=8# passes through the points #(8,1)# and #(0,3)#. The slope of the line that passes through these points is #f'(8)=(3-1)/(0-8)=2/(-8)=-1/4#.

Isaiah Allen · Answer

undefined To find f'(8), we can use the fact that the line tangent to the graph of function f at the point (8,1) intersects the y-axis at y=3. The slope of this tangent line is equal to the derivative of f at x=8, which is f'(8). Since the line intersects the y-axis at y=3, we know that the y-coordinate of the point (8,1) is 3 units below the y-intercept of the tangent line. Therefore, the slope of the tangent line is equal to the change in y divided by the change in x, which is (1-3)/(8-0) = -2/8 = -1/4. Hence, f'(8) = -1/4.