Question

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?

Answer 1

`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