Question

What is the slope of the tangent line for the function f(x)=2x^2-5x+1 at the point (1,-2)

A. -1
B.-9
C. 1
D. 0
E.9

Answer 1

This is given by the limit

`\displaystyle \lim_(x\to1) (f(x) - f(1))/(x - 1) = \lim_(x\to1) ((2x^2 - 5x+1) - (-2))/(x - 1) = \lim_(x\to1) (2x^2 - 5x + 3)/(x - 1)`

Observe that
`2x^2-5x+3=0` when
`x=1`, and factorizing gives

`2x^2 - 5x + 3 = (x - 1) (2x - 3)`

so that

`\displaystyle \lim_(x\to1) (f(x) - f(1))/(x - 1) = \lim_(x\to1) (2x-3) = 2\cdot1 - 3 = \boxed{-1}`

(A)