What is the tangent of f(x)=sqrt(x^2-x+13),\at x=4 ?

The tangent of f(x)=sqrt(x^2-x+13),\at x=4 is y= 7/10 x+11/5

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tangent of f(x)=sqrt(x^2-x+13),\at x=4

tangent of

f (x) = x^{2} - x + 13, at x = 4

y = \frac{7}{10} x + \frac{11}{5}

Solution steps

Find the tangent point:

(4, 5)

Find the slope of

f (x) = x^{2} - x + 13 : \frac{df ( x )}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 13}

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 = \frac{7}{10}

Find the line with slope m=

\frac{7}{10}

and passing through

(4, 5) : y = \frac{7}{10} x + \frac{11}{5}

y = \frac{7}{10} x + \frac{11}{5}

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

\frac{r}{r ^{2} + 3}

8 sin (t)

k = 1 \sum \infty \frac{k ^{3}}{5 ^{k}}

\int \frac{7 sin ( x ) + 2 sec ( x )}{tan ( x )} d x

What is the tangent of f(x)=sqrt(x^2-x+13),\at x=4 ?
The tangent of f(x)=sqrt(x^2-x+13),\at x=4 is y= 7/10 x+11/5