How do you find the slope of the tangent line to the graph at given point and give an equation of the tangent line #f(x)=3/2x# at (1, 3/2)?

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Answer 1 · 2017-07-26T13:07:08

The answer may seem strange when compared to what we know about tangent lines to circles and other curves. Please see below.

This function is a linear function. Its graph is the straight line #y = 3/2x#.

The tangent line can be described as the limit, as a second point approaches #(1,3/2)#, of the slope of secant line.

For every point on the graph, the secant line is the same as the graph of the original function, #y = 3/2x#.

Therefore, the tangent line is the same, #y = 3/2x#.

If we want the details:

The slope of the tangent line is given by

#lim_(hrarr0)(f(1+h)-f(1))/h = lim_(hrarr0)(3/2(1+h)-3/2(1))/h#

# = lim_(hrarr0)(3/2cancel(h))/cancel(h)#

# = 3/2#

The line through the point #(x_1,y_1)# with slop #m=3/2# can be written

#y-y_1 = m(x-x_1)#

So we have

#y-3/2 = 3/2(x-1)# Solving for #y# to get slope intercept form, we get

#y = 3/2x#