What is the tangent of f(x)=-(9x^2)/2+8x-22 ?

The tangent of f(x)=-(9x^2)/2+8x-22 is y=(-9a_{0}+8)x+9a_{0}^2-(9a_{0}^2)/2-22

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tangent of f(x)=-(9x^2)/2+8x-22

tangent of

f (x) = - \frac{9 x ^{2}}{2} + 8 x - 22

y = (- 9 a_{0} + 8) x + 9 a_{0}^{2} - \frac{9 a _{0}^{2}}{2} - 22

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{9 a _{0}^{2}}{2} + 8 a_{0} - 22)

Find the slope of

f (x) = - \frac{9 x ^{2}}{2} + 8 x - 22 : \frac{df ( x )}{d x} = - 9 x + 8

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 9 a_{0} + 8

Find the line with slope m=

- 9 a_{0} + 8

and passing through

(a_{0}, - \frac{9 a _{0}^{2}}{2} + 8 a_{0} - 22) : y = (- 9 a_{0} + 8) x + 9 a_{0}^{2} - \frac{9 a _{0}^{2}}{2} - 22

y = (- 9 a_{0} + 8) x + 9 a_{0}^{2} - \frac{9 a _{0}^{2}}{2} - 22

\frac{d h}{d t} = - h^{\frac{1}{3}}

\int (x^{2})^{23} d (x^{2})^{2}

x e^{e}

y = - 2 x

5 e^{- x^{2}}

What is the tangent of f(x)=-(9x^2)/2+8x-22 ?
The tangent of f(x)=-(9x^2)/2+8x-22 is y=(-9a_{0}+8)x+9a_{0}^2-(9a_{0}^2)/2-22