What is the tangent of f(x)=3-4x^2,-61,\at x=4 ?

The tangent of f(x)=3-4x^2,-61,\at x=4 is y=-32x+67

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tangent of f(x)=3-4x^2,-61,\at x=4

tangent of

f (x) = 3 - 4 x^{2}, - 61, at x = 4

y = - 32 x + 67

Solution steps

Find the tangent point:

(4, - 61)

Find the slope of

f (x) = 3 - 4 x^{2} : \frac{df ( x )}{d x} = - 8 x

Compute the slope of

f (x) = 3 - 4 x^{2}

x = 4 : m = - 32

Find the line with slope m=

- 32

and passing through

(4, - 61) : y = - 32 x + 67

y = - 32 x + 67

n = 1 \sum \infty arctan (n)

\int 2 x^{3} + 4 d x

\frac{20}{s ( s ^{2} + 2 s + 5 )}

x \to 8 - lim (\frac{x + 7}{x + 8})

\int \frac{1}{4 - ( x - 1 ) ^{2}} d x

What is the tangent of f(x)=3-4x^2,-61,\at x=4 ?
The tangent of f(x)=3-4x^2,-61,\at x=4 is y=-32x+67