What is the tangent of y=3x^2-5x+9 ?

The tangent of y=3x^2-5x+9 is y=(6a_{0}-5)x-3a_{0}^2+9

Popular >

tangent of y=3x^2-5x+9

tangent of

y = 3 x^{2} - 5 x + 9

y = (6 a_{0} - 5) x - 3 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 5 a_{0} + 9)

Find the slope of

y = 3 x^{2} - 5 x + 9 : \frac{d y}{d x} = 6 x - 5

Compute the slope of

y = 3 x^{2} - 5 x + 9

x = a_{0} : m = 6 a_{0} - 5

Find the line with slope m=

6 a_{0} - 5

and passing through

(a_{0}, 3 a_{0}^{2} - 5 a_{0} + 9) : y = (6 a_{0} - 5) x - 3 a_{0}^{2} + 9

y = (6 a_{0} - 5) x - 3 a_{0}^{2} + 9

y^{^{''}} + 9 y = 3 sec (3 x)

5 x e^{2 x}

\frac{d}{d x} (\frac{x ^{2}}{4 - x ^{2} - y ^{2}})

x \to 0 lim (e^{x} + 2 x)

k = 3 \sum \infty \frac{1}{ln ( k )}

What is the tangent of y=3x^2-5x+9 ?
The tangent of y=3x^2-5x+9 is y=(6a_{0}-5)x-3a_{0}^2+9