What is the tangent of f(x)=-8x^2+5x,\at x=-1 ?

The tangent of f(x)=-8x^2+5x,\at x=-1 is y=21x+8

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tangent of f(x)=-8x^2+5x,\at x=-1

tangent of

f (x) = - 8 x^{2} + 5 x, at x = - 1

y = 21 x + 8

Solution steps

Find the tangent point:

(- 1, - 13)

Find the slope of

f (x) = - 8 x^{2} + 5 x : \frac{df ( x )}{d x} = - 16 x + 5

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 21

Find the line with slope m=

21

and passing through

(- 1, - 13) : y = 21 x + 8

y = 21 x + 8

What is the tangent of f(x)=-8x^2+5x,\at x=-1 ?
The tangent of f(x)=-8x^2+5x,\at x=-1 is y=21x+8