If the tangent line to #y = f(x)# at #(4,3)# passes through the point #(0,2)#, Find #f(4)# and #f'(4)#? An explanation would also be very helpful.

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Answer 1 · 2017-03-04T20:59:30

#f(4) = 3#

#f'(4) = 1/4#

The question gives you #f(4)# already, because the point #(4,3)# is given. When #x# is #4#, #[y = f(x) = ]f(4)# is #3#.

We can find #f'(4)# by finding the gradient at the point #f(4)#, which we can do because we know the tangent touches both #(4,3)# and #(0,2)#.

The gradient of a line is given by rise over run, or the change in #y# divided by the change in #x#, or, mathematically,

#m = (y_2-y_1)/(x_2-x_1)#

We know two points on the graph in the question, so effectively we know the two values we need for #y# and #x# each. Say that

#(0,2) -> x_1 = 0, y_1 = 2#

#(4,3) -> x_2 = 4, y_2 = 3#

so

#m = (3-2)/(4-0) = 1/4#

which is the gradient.