How do I find the equation of the tangent to the graph of f(x)=Sqrt x at the point x=9.
I know the gradient of the tangent which is 1/6.
But I have no idea how to do this.
Any help will be appreciated.
THANKS
Hello Awsom GuyAs you say, $\displaystyle f'(9) = \frac{1}{2\sqrt9}=\frac16$.
You now need $\displaystyle f(9) = \sqrt9 = 3$
So the tangent is at the point $\displaystyle (9,3)$.
Now use $\displaystyle y - y_1 = m(x-x_1)$, the equation of the line with gradient $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$.
When simplified, this comes to $\displaystyle 6y = x+9$.
Can you fill in the details?
Grandad
Hello sjaraThe value of $\displaystyle m$ is the gradient of the curve at the point where $\displaystyle x = 9$. We get this by differentiating $\displaystyle f(x)$, and then plugging in that value, $\displaystyle x = 6$.
$\displaystyle f(x) = \sqrt{x}=x^{\frac12}$
$\displaystyle \Rightarrow f'(x)=\tfrac12x^{-\frac12}=\frac{1}{2\sqrt{x}}$
So when $\displaystyle x = 9, f'(9)= \frac{1}{2\sqrt{9}}=\frac16$
And this is the value of $\displaystyle m$, the gradient of the tangent when $\displaystyle x = 9$.
Grandad