The equation of the tangent to y = \(\rm\sqrt{10 − 3x}\) at the point where x = 3 is

To find the equation of the tangent to the curve y = \(\rm\sqrt{(10 − 3x)}\) at the point where x = 3, we need to find the slope of the tangent and a point on the tangent line.

First, let's find the slope of the tangent at the point (3, \(\rm\sqrt{(10 − 3\times 3)}\)) = (3, 1).

To find the slope, we differentiate the given function y = √(10 - 3x) with respect to x:

\(\frac{dy}{dx}= \frac{d}{dx}(\rm\sqrt{(10 − 3x)}) \\ \Rightarrow \frac{dy}{dx}=\frac{-3}{2\times\sqrt{(10 − 3x)}}\)

Now, substitute x = 3 into the derived expression:

\(\Rightarrow \frac{dy}{dx}=\frac{-3}{2\times\sqrt{(10 − 3\times3)}}=\frac{-3}{2}\)

So, the slope of the tangent at the point (3, 1) is -3/2.

Next, we can use the point-slope form of a linear equation to find the equation of the tangent:

y - y1 = m(x - x1)

where (x1, y1) is the point (3, 1) and m is the slope, -3/2.

Substituting the values:

y - 1 = (-3/2)(x - 3)

Simplifying:

y - 1 = (-3/2)x + 9/2
y = (-3/2)x + 9/2 + 1
y = (-3/2)x + 9/2 + 2/2
y = (-3/2)x + 11/2

Multiplying both sides of the equation by 2 to eliminate fractions:

2y = -3x + 11

Finally, rearrange the equation to match the given options:

3x + 2y = 11

Therefore, the correct answer is: 3x + 2y = 11