What is the tangent of f(x)=-x^2+16x+63,\at x=2 ?

The tangent of f(x)=-x^2+16x+63,\at x=2 is y=12x+67

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tangent of f(x)=-x^2+16x+63,\at x=2

tangent of

f (x) = - x^{2} + 16 x + 63, at x = 2

y = 12 x + 67

Solution steps

Find the tangent point:

(2, 91)

Find the slope of

f (x) = - x^{2} + 16 x + 63 : \frac{df ( x )}{d x} = - 2 x + 16

Compute the slope of

f (x) = - x^{2} + 16 x + 63

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 91) : y = 12 x + 67

y = 12 x + 67

\int ln (4 x) d x

\int_{1}^{12} (0.5)^{x} d x

\frac{d}{d x} (\frac{1}{1 - cos ( x )})

\int \frac{2 x}{x ^{2} - 4 x + 13} d x

\frac{d}{d x} (x e^{x} cot (x))

What is the tangent of f(x)=-x^2+16x+63,\at x=2 ?
The tangent of f(x)=-x^2+16x+63,\at x=2 is y=12x+67