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

The tangent of f(x)=4x^2-4x+1,\at x=4 is y=28x-63

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

tangent of

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

y = 28 x - 63

Solution steps

Find the tangent point:

(4, 49)

Find the slope of

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

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 = 28

Find the line with slope m=

28

and passing through

(4, 49) : y = 28 x - 63

y = 28 x - 63

\frac{\partial}{\partial x} (y \cdot e^{x})

f (x) = \frac{x}{x + 3}

n = 1 \sum \infty \frac{( - 1 ) ^{n + 1} 2 n}{n + 1}

\int_{- \frac{π}{4}}^{0} 6 sec^{3} (x) d x

x \to 2 - lim (\frac{∣ x - 2 ∣}{x - 2})

What is the tangent of f(x)=4x^2-4x+1,\at x=4 ?
The tangent of f(x)=4x^2-4x+1,\at x=4 is y=28x-63