Helppppppppp !!!!!!!!!!!!

Question

Find the equation of the tangent line to the function 𝑓(𝑥)=4√x−6 at the point (16,10). Provide your answer in slope–intercept form of a linear equation, 𝑦=𝑚𝑥+𝑏, where 𝑚 is the slope and 𝑏 is the 𝑦- intercept. Express 𝑚 and 𝑏 as exact numbers.

Answer 1

First we need to find the slope of the tangent line to the curve at (16, 10). To do this, we need the derivative of the function:

f'(x) = 2 * x^-1/2

f'(16) = 2 * 16^-1/2 = 1/2

Then we can start with the point/slope form of a line with the slope and the given point:

(y - 10) = 1/2 * (x - 16)

And simplify:

y = x/2 + 2

Answer 2

Hi Anjela R.,

The tangent line matches the slope of the function at the point of tangency (what else could it do, after all?).

So all you need to do is to find that slope. What is slope of a function? It's the value of the first derivative of the function with respect to x.

So: f(x) = 4* (x)^0.5 - 6

f ' (x) = d (4* (x)^0.5 - 6)/dx

So how do you take the derivative of a function that has several parts to it?

the - 6 disappears!

the 4* part stays, but then the d() propagates in to the (x)^0.5 .

So how do you take the derivative of an power of x?

The derivative of x^n is nx^(n-1) . So apply that: d(x^0.5)/dx = 0.5*x^(-0.5) .

Then, plug into the form for a point and a slope: (y - y1) = m (x - x1), and reduce that to the form you want.

You know how to manipulate equations algebraically, right?

-- Cheers, --Mr. d.