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)?

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Answer 1 · 2016-05-24T21:04:38

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#

Answer 2 · 2016-05-26T02:42:58

#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#.