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

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

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

tangent of

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

y = 5 x + 1

Solution steps

Find the slope of

y = 4 x^{2} + 5 x + 1 : \frac{d y}{d x} = 8 x + 5

Compute the slope of

y = 4 x^{2} + 5 x + 1

(0, 1) : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x + 1

\frac{d}{d x} (lo g_{10} (lo g_{10} (x)))

\int sin^{2} (θ) cos (4 θ) d θ

n = 1 \sum \infty \frac{3 n !}{n ^{n}}

\int_{0}^{\frac{π}{2}} sin (2 x) d x

3 t^{4} - 2 t^{2} + 1

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