What is the tangent of f(x)=4x^2+5x+1,(0,1) ?

The tangent of f(x)=4x^2+5x+1,(0,1) is y=5x+1

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tangent of f(x)=4x^2+5x+1,(0,1)

tangent of

f (x) = 4 x^{2} + 5 x + 1, (0, 1)

y = 5 x + 1

Solution steps

Find the slope of

f (x) = 4 x^{2} + 5 x + 1 : \frac{df ( x )}{d x} = 8 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 = 5

Find the line with slope m=

5

and passing through

(0, 1) : y = 5 x + 1

y = 5 x + 1

y = \frac{4 - 3 x}{3 x ^{2} + x}

630 - 4.9 t^{2}

x \to \frac{π}{2} lim (8 sec (x))

y^{^{'}} = y (x y^{4} + 3)

y^{^{'}} - \frac{2 y}{x} = 1

What is the tangent of f(x)=4x^2+5x+1,(0,1) ?
The tangent of f(x)=4x^2+5x+1,(0,1) is y=5x+1