What is the tangent of 8x^2+3x-7 ?

The tangent of 8x^2+3x-7 is y=(16a_{0}+3)x-8a_{0}^2-7

tangent of 8x^2+3x-7

tangent of

8 x^{2} + 3 x - 7

y = (16 a_{0} + 3) x - 8 a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} + 3 a_{0} - 7)

Find the slope of

y = 8 x^{2} + 3 x - 7 : \frac{d y}{d x} = 16 x + 3

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 = 16 a_{0} + 3

Find the line with slope m=

16 a_{0} + 3

and passing through

(a_{0}, 8 a_{0}^{2} + 3 a_{0} - 7) : y = (16 a_{0} + 3) x - 8 a_{0}^{2} - 7

y = (16 a_{0} + 3) x - 8 a_{0}^{2} - 7

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

d er i v a t i v e 8 (e^{0} cos (0) - e^{0} sin (0))

t an g e n t \frac{x ^{2}}{x + 9}

\int \frac{3 - 2 x}{5 x ^{2} + 7} d x

\int_{- 1}^{3} (x^{3} - 2) d x

What is the tangent of 8x^2+3x-7 ?
The tangent of 8x^2+3x-7 is y=(16a_{0}+3)x-8a_{0}^2-7