What is the tangent of y=4x^2-3x+3,(1,4) ?

The tangent of y=4x^2-3x+3,(1,4) is y=5x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=4x^2-3x+3,(1,4)

tangent of

y = 4 x^{2} - 3 x + 3, (1, 4)

y = 5 x - 1

Solution steps

Find the slope of

y = 4 x^{2} - 3 x + 3 : \frac{d y}{d x} = 8 x - 3

Compute the slope of

y = 4 x^{2} - 3 x + 3

(1, 4) : m = 5

Find the line with slope m=

5

and passing through

(1, 4) : y = 5 x - 1

y = 5 x - 1

\int x^{3} - 3 d x

\frac{d}{d θ} (e^{t a n (θ)})

\int_{0}^{1} \frac{y}{e ^{5 y}} d y

\int (2 x x - \frac{1}{x}) d x

3 y^{^{''}} - 11 y^{^{'}} + 10 y = 0, y (0) = 8, y^{^{'}} (0) = - 4

What is the tangent of y=4x^2-3x+3,(1,4) ?
The tangent of y=4x^2-3x+3,(1,4) is y=5x-1