What is the tangent of f(x)=4x^2+3x+6,(-4,58) ?

The tangent of f(x)=4x^2+3x+6,(-4,58) is y=-29x-58

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tangent of f(x)=4x^2+3x+6,(-4,58)

tangent of

f (x) = 4 x^{2} + 3 x + 6, (- 4, 58)

y = - 29 x - 58

Solution steps

Find the slope of

f (x) = 4 x^{2} + 3 x + 6 : \frac{df ( x )}{d x} = 8 x + 3

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 = - 29

Find the line with slope m=

- 29

and passing through

(- 4, 58) : y = - 29 x - 58

y = - 29 x - 58

\frac{d}{d x} (cos (a x) a)

f (x) = I n (2 x + 1)

\int sec^{4} (x) tan^{2} (x) d x

\frac{d}{d x} ((x + 2 x) e^{x})

\int \frac{1}{x ^{2} - 3 x + 2} d x

What is the tangent of f(x)=4x^2+3x+6,(-4,58) ?
The tangent of f(x)=4x^2+3x+6,(-4,58) is y=-29x-58