What is the tangent of y=-9/2 x^2+3x+3,(-1,-9/2) ?

The tangent of y=-9/2 x^2+3x+3,(-1,-9/2) is y=12x+15/2

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tangent of y=-9/2 x^2+3x+3,(-1,-9/2)

tangent of

y = - \frac{9}{2} x^{2} + 3 x + 3, (- 1, - \frac{9}{2})

y = 12 x + \frac{15}{2}

Solution steps

Find the slope of

y = - \frac{9}{2} x^{2} + 3 x + 3 : \frac{d y}{d x} = - 9 x + 3

Compute the slope of

y = - \frac{9}{2} x^{2} + 3 x + 3

(- 1, - \frac{9}{2}) : m = 12

Find the line with slope m=

12

and passing through

(- 1, - \frac{9}{2}) : y = 12 x + \frac{15}{2}

y = 12 x + \frac{15}{2}

x \to 2 lim (x^{2} - 2)

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

\frac{z ( z + 1 )}{z ^{2}}

x \to π + lim (h (x))

\frac{\partial}{\partial x} (4 π^{2} L T^{- 2})

What is the tangent of y=-9/2 x^2+3x+3,(-1,-9/2) ?
The tangent of y=-9/2 x^2+3x+3,(-1,-9/2) is y=12x+15/2