How do you write an equation of the line tangent to the graph of #f(x)=7-15x+9x^2-x^3# at x=2?

1 Answer
Sep 26, 2015

#y=9x-13=># equation of tangent in slope-intercept form

Explanation:

#f(x)=7−15x+9x^2−x^3=># evaluate #f(2):#
#f(2)=7-15*2+9*2^2-2^3#
#f(2)=7-30+36-8=5=># tangent at point:#(2, 5)#
The slope of the tangent line to graph of a function at a point is equal to first derivative of the function evaluated at that point:
#f'(x)=-15+18x-3x^2=-3(x^2-6x+5)#
#f'(2)=-3(4-12+5)=9#
Slope#=m=9# , point#(x_1, y_1)=(2, 5)# , using point-slope form:
#y-y_1=m(x-x_1):#
#y-5=9(x-2)=># equation of tangent in point-slope form
#y=9x-13=># equation of tangent in slope-intercept form