What is the tangent of f(x)=(x-2)(2x+3),\at x=0 ?

The tangent of f(x)=(x-2)(2x+3),\at x=0 is y=-x-6

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tangent of f(x)=(x-2)(2x+3),\at x=0

tangent of

f (x) = (x - 2) (2 x + 3), at x = 0

y = - x - 6

Solution steps

Find the tangent point:

(0, - 6)

Find the slope of

f (x) = (x - 2) (2 x + 3) : \frac{df ( x )}{d x} = 4 x - 1

Compute the slope of

f (x) = (x - 2) (2 x + 3)

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

(0, - 6) : y = - x - 6

y = - x - 6

\int (x - 5) \cdot sin (x) d x

\int \frac{sin ( x ) tan ( x )}{9 + sec ^{2} ( x )} d x

t^{2} y^{^{''}} + 21 t y^{^{'}} + 100 y = 0

f (x) = sin (x) + \frac{6}{7} cot (x)

x \to 2 lim (5 a + a x)

What is the tangent of f(x)=(x-2)(2x+3),\at x=0 ?
The tangent of f(x)=(x-2)(2x+3),\at x=0 is y=-x-6