What is the tangent of f(x)=2x^2+10x+9,\at x=-3 ?

The tangent of f(x)=2x^2+10x+9,\at x=-3 is y=-2x-9

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tangent of f(x)=2x^2+10x+9,\at x=-3

tangent of

f (x) = 2 x^{2} + 10 x + 9, at x = - 3

y = - 2 x - 9

Solution steps

Find the tangent point:

(- 3, - 3)

Find the slope of

f (x) = 2 x^{2} + 10 x + 9 : \frac{df ( x )}{d x} = 4 x + 10

Compute the slope of

f (x) = 2 x^{2} + 10 x + 9

x = - 3 : m = - 2

Find the line with slope m=

- 2

and passing through

(- 3, - 3) : y = - 2 x - 9

y = - 2 x - 9

\frac{( s + 5 )}{( s + 6 ) ( s + 10 )}

n = 0 \sum \infty \frac{n}{n ^{2} + 4}

\int_{0}^{1} 9

\int 4 x^{9} e^{- x^{5}} d x

y = 2 x, y = 5 x^{2}

What is the tangent of f(x)=2x^2+10x+9,\at x=-3 ?
The tangent of f(x)=2x^2+10x+9,\at x=-3 is y=-2x-9