What is the tangent of f(x)=-x^2-3x+4,\at x=-3 ?

The tangent of f(x)=-x^2-3x+4,\at x=-3 is y=3x+13

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tangent of f(x)=-x^2-3x+4,\at x=-3

tangent of

f (x) = - x^{2} - 3 x + 4, at x = - 3

y = 3 x + 13

Solution steps

Find the tangent point:

(- 3, 4)

Find the slope of

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

Compute the slope of

f (x) = - x^{2} - 3 x + 4

x = - 3 : m = 3

Find the line with slope m=

3

and passing through

(- 3, 4) : y = 3 x + 13

y = 3 x + 13

\int \frac{x ^{2} - 144}{x} d x

\int cos (y) cos (z) d y

\int_{2}^{3} (\frac{4}{x ^{2}} + 1) d x

\int \frac{- 7}{( x ^{2} + 7 ) ^{\frac{3}{2}}} d x

n = 8 \sum \infty \frac{n}{ln ( n )}

What is the tangent of f(x)=-x^2-3x+4,\at x=-3 ?
The tangent of f(x)=-x^2-3x+4,\at x=-3 is y=3x+13