What is the tangent of y=5-9x^2 ?

The tangent of y=5-9x^2 is y=-18a_{0}x+5+9a_{0}^2

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tangent of y=5-9x^2

tangent of

y = 5 - 9 x^{2}

y = - 18 a_{0} x + 5 + 9 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 - 9 a_{0}^{2})

Find the slope of

y = 5 - 9 x^{2} : \frac{d y}{d x} = - 18 x

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 = - 18 a_{0}

Find the line with slope m=

- 18 a_{0}

and passing through

(a_{0}, 5 - 9 a_{0}^{2}) : y = - 18 a_{0} x + 5 + 9 a_{0}^{2}

y = - 18 a_{0} x + 5 + 9 a_{0}^{2}

\int \frac{1}{16 x ^{4} - 1} d x

\frac{d}{d x} (e^{r (t, x) t})

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

y = ((25 - x^{2})^{\frac{1}{2}}), y = \frac{4}{3} x, x = 5

(\frac{1 + x ^{2}}{1 - x ^{2}})^{17}

What is the tangent of y=5-9x^2 ?
The tangent of y=5-9x^2 is y=-18a_{0}x+5+9a_{0}^2