What is the tangent of f(x)=sqrt(x^2+3x),(1,2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x^2+3x),(1,2)

tangent of

f (x) = x^{2} + 3 x, (1, 2)

y = \frac{5}{4} x + \frac{3}{4}

Solution steps

Find the slope of

f (x) = x^{2} + 3 x : \frac{df ( x )}{d x} = \frac{2 x + 3}{2 x ^{2} + 3 x}

Compute the slope of

f (x) = x^{2} + 3 x

(1, 2) : m = \frac{5}{4}

Find the line with slope m=

\frac{5}{4}

and passing through

(1, 2) : y = \frac{5}{4} x + \frac{3}{4}

y = \frac{5}{4} x + \frac{3}{4}

\int \frac{9 - 9 x}{1 - x} d x

y = x, at x = 4

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

y^{^{'''}} + 4 y^{^{''}} + 4 y^{^{'}} = 8 x

\int x^{4} e^{\frac{x}{2}} d x

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