Question

Etermine an equation of the line that is tangent to the graph of f(x) =`sqrt(x+1)` and parallel to `x-6y+4=0`?

Answer 1

The equation is `y = 1/6x + 5/3`

Explanation:

We see that the line represented by `x - 6y + 4 = 0` can be converted to slope intercept form .

`x + 4 = 6y`
`1/6x + 2/3 = y`

Therefore the line has slope `1/6`. Since the tangent is parallel to this line, it will have an equal slope.

Now we find the derivative, because it will give us the slope of the tangent. Use the chain rule.

`f'(x) = 1/(2sqrt(x + 1))`

We are looking for when

`1/6 = 1/(2sqrt(x + 1))`

`2/6sqrt(x + 1) = 1`

`sqrt(x + 1) = 3`

`x + 1 = 9`

`x = 8`

It follows that `f(8) = sqrt(8 + 1) = 3`.

The equation of the tangent is therefore

`y - y_1 = m(x - x_1)`

`y - 3 = 1/6(x- 8)`

`y = 1/6x - 4/3 + 3`

`y = 1/6x + 5/3`

We can confirm graphically.

The green line is the tangent line, the blue graph is `f(x)` and the red line is the line parallel to the tangent. Clearly our answer is correct because the tangent line is parallel to the initial line is truly tangent to the square root graph (touches at one point).

Hopefully this helps!

Answer 2

See solution below

Explanation:

We know that the general line equation is `y=mx+b` where `m` is the slope, and `b` is the line intercept with `y` axis

By other hand, we know that `f'(x_0)` is the slope of a tangent line at `x_0`

In our case we don't know what is the value of `x_0`

But the parallel line given `x-6y+4=0` is the same that `y=1/6x+4/6` (transposing terms). So the slope of tangent line is the same because both are parallels. That's to say `m=1/6`

We know now that searched equation is `y=1/6x+b`. Lets determine `b`

At the tangency point, the value of `f(x_0)=y` given by the line equation because graph and line are tangents in that point. Thus we have

`f'(x_0)= 1/(2sqrt(1+x_0))=1/6`. Transposing terms we have `x_0=8`

Thus `f(8)=3=1/6·8+b`. Transposing, we find `b=5/3`

So, the tangent line requested is `y=1/6x+5/3` at the point `x_0=8`