What is the tangent of f(x)=-9/2 x^2+7x+7,\at x=-1 ?

The tangent of f(x)=-9/2 x^2+7x+7,\at x=-1 is y=16x+23/2

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tangent of f(x)=-9/2 x^2+7x+7,\at x=-1

tangent of

f (x) = - \frac{9}{2} x^{2} + 7 x + 7, at x = - 1

y = 16 x + \frac{23}{2}

Solution steps

Find the tangent point:

(- 1, - \frac{9}{2})

Find the slope of

f (x) = - \frac{9}{2} x^{2} + 7 x + 7 : \frac{df ( x )}{d x} = - 9 x + 7

Compute the slope of

f (x) = - \frac{9}{2} x^{2} + 7 x + 7

x = - 1 : m = 16

Find the line with slope m=

16

and passing through

(- 1, - \frac{9}{2}) : y = 16 x + \frac{23}{2}

y = 16 x + \frac{23}{2}

16 - x, 12, 16

\frac{d}{d x} ((3 + 4 x - x^{2})^{\frac{1}{2}})

y^{^{''}} + 4 y = 2 t^{2}

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

\frac{d}{d x} (lo g_{e} (tan (x)))

What is the tangent of f(x)=-9/2 x^2+7x+7,\at x=-1 ?
The tangent of f(x)=-9/2 x^2+7x+7,\at x=-1 is y=16x+23/2