What is the tangent of f(x)=-x^2+4x,\at x=1 ?

The tangent of f(x)=-x^2+4x,\at x=1 is y=2x+1

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tangent of f(x)=-x^2+4x,\at x=1

tangent of

f (x) = - x^{2} + 4 x, at x = 1

y = 2 x + 1

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

f (x) = - x^{2} + 4 x

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 3) : y = 2 x + 1

y = 2 x + 1

\frac{x d y}{d x} + 5 y = x^{3} - x

\int_{0}^{1} \frac{e ^{x} + 1}{e ^{x} + x} d x

y^{^{'}} = y sin (x), y (0) = 1

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{2} e^{3 t}

\frac{\partial}{\partial x} (tan (y))

What is the tangent of f(x)=-x^2+4x,\at x=1 ?
The tangent of f(x)=-x^2+4x,\at x=1 is y=2x+1