What is the tangent of f(x)=-4x^2-5x+2 ?

The tangent of f(x)=-4x^2-5x+2 is y=(-8a_{0}-5)x+4a_{0}^2+2

Popular >

tangent of f(x)=-4x^2-5x+2

tangent of

f (x) = - 4 x^{2} - 5 x + 2

y = (- 8 a_{0} - 5) x + 4 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 4 a_{0}^{2} - 5 a_{0} + 2)

Find the slope of

f (x) = - 4 x^{2} - 5 x + 2 : \frac{df ( x )}{d x} = - 8 x - 5

Compute the slope of

f (x) = - 4 x^{2} - 5 x + 2

x = a_{0} : m = - 8 a_{0} - 5

Find the line with slope m=

- 8 a_{0} - 5

and passing through

(a_{0}, - 4 a_{0}^{2} - 5 a_{0} + 2) : y = (- 8 a_{0} - 5) x + 4 a_{0}^{2} + 2

y = (- 8 a_{0} - 5) x + 4 a_{0}^{2} + 2

x \to 4 lim (\frac{x - 2}{x - 4})

\int_{0}^{π} x cos (3 x) d x

t^{2} y^{^{''}} + 9 t y^{^{'}} + 7 y = 0

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

\int \frac{y ^{2} + 2 y + 1}{( y ^{2} + 1 ) ^{2}} d y

What is the tangent of f(x)=-4x^2-5x+2 ?
The tangent of f(x)=-4x^2-5x+2 is y=(-8a_{0}-5)x+4a_{0}^2+2