What is the tangent of f(x)= 11/2-x^2,\at x=-2 ?

The tangent of f(x)= 11/2-x^2,\at x=-2 is y=4x+19/2

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tangent of f(x)= 11/2-x^2,\at x=-2

tangent of

f (x) = \frac{11}{2} - x^{2}, at x = - 2

y = 4 x + \frac{19}{2}

Solution steps

Find the tangent point:

(- 2, \frac{3}{2})

Find the slope of

f (x) = \frac{11}{2} - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = \frac{11}{2} - x^{2}

x = - 2 : m = 4

Find the line with slope m=

4

and passing through

(- 2, \frac{3}{2}) : y = 4 x + \frac{19}{2}

y = 4 x + \frac{19}{2}

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k = 0 \sum \infty (\frac{1}{2})^{2 k + 1}

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What is the tangent of f(x)= 11/2-x^2,\at x=-2 ?
The tangent of f(x)= 11/2-x^2,\at x=-2 is y=4x+19/2