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

The tangent of f(x)=8x^2+5x,\at 53 is y=(16a_{0}+5)x-8a_{0}^2

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tangent of f(x)=8x^2+5x,\at 53

tangent of

f (x) = 8 x^{2} + 5 x, at 53

y = (16 a_{0} + 5) x - 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} + 5 a_{0})

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 = 16 a_{0} + 5

Find the line with slope m=

16 a_{0} + 5

and passing through

(a_{0}, 8 a_{0}^{2} + 5 a_{0}) : y = (16 a_{0} + 5) x - 8 a_{0}^{2}

y = (16 a_{0} + 5) x - 8 a_{0}^{2}

e^{20 t + 20}

\frac{\partial}{\partial y} (∣ x ∣ y + 2 x)

\frac{d}{d x} (3 (1 - x^{2}))

\frac{d}{d x} (- 8 sin (x))

x \to 2 lim (\frac{( 2 x ^{2} + 1 )}{( 3 x - 2 )})

What is the tangent of f(x)=8x^2+5x,\at 53 ?
The tangent of f(x)=8x^2+5x,\at 53 is y=(16a_{0}+5)x-8a_{0}^2