What is the tangent of f(x)=(1+2x)(6-5x),\at x=1 ?

The tangent of f(x)=(1+2x)(6-5x),\at x=1 is y=-13x+16

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tangent of f(x)=(1+2x)(6-5x),\at x=1

tangent of

f (x) = (1 + 2 x) (6 - 5 x), at x = 1

y = - 13 x + 16

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = (1 + 2 x) (6 - 5 x) : \frac{df ( x )}{d x} = - 20 x + 7

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 = - 13

Find the line with slope m=

- 13

and passing through

(1, 3) : y = - 13 x + 16

y = - 13 x + 16

\frac{\partial}{\partial x} (2 ln (x y))

\int x^{2} arcsin (x) d x

\int \frac{( ln ( x ) ) ^{28}}{x} d x

\frac{d}{d x} (arcsin (3 x + 1))

\frac{\partial}{\partial z} (x y z^{2} tan (y t))

What is the tangent of f(x)=(1+2x)(6-5x),\at x=1 ?
The tangent of f(x)=(1+2x)(6-5x),\at x=1 is y=-13x+16