What is the tangent of f(x)=8x^2+5x,\at x=3 ?

The tangent of f(x)=8x^2+5x,\at x=3 is y=53x-72

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tangent of f(x)=8x^2+5x,\at x=3

tangent of

f (x) = 8 x^{2} + 5 x, at x = 3

y = 53 x - 72

Solution steps

Find the tangent point:

(3, 87)

Find the slope of

f (x) = 8 x^{2} + 5 x : \frac{df ( x )}{d x} = 16 x + 5

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 = 53

Find the line with slope m=

53

and passing through

(3, 87) : y = 53 x - 72

y = 53 x - 72

x \to 1.001 lim (x^{2} - 2 x - 1)

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

\int e^{- 4 x} \cdot x d x

d er i v a t i v e 2 x^{2} + 15

t an g e n t y = sec (x), (\frac{π}{6}, 2 \frac{3}{3})

What is the tangent of f(x)=8x^2+5x,\at x=3 ?
The tangent of f(x)=8x^2+5x,\at x=3 is y=53x-72