How do you find the equation of a line tangent to the function #y=x+4/sqrtx# at (1,5)?

Question

Evelyn Agard · Accepted Answer

#y=-x+6# #y=x+4/sqrtx# #= x+4x^(-1/2)# #y' = d/dx(x+4x^(-1/2))# #=1-2x^(-3/2) = 1-2/sqrt(x^3)# The equation of a straight line through #(x_1, y_1)# is: #(y-y_1) = m(x-x_1)# Where m is the slope the line The slope of #y# at #(1, 5)# is #y'(1) = 1-2/1 = -1# Hence the slope of the tangent to #y# at #(1, 5)# is: #(y-5) = -1(x-1)# #y=-x+6# graph{(y-x-4/sqrtx)(y+x-6)=0 [-7.04, 12.96, -1.04, 8.96]}

William Abdalla · Answer

undefined To find the equation of a line tangent to a function at a given point, we need to find the derivative of the function and evaluate it at the given point.

First, let's find the derivative of the function y = (x + 4) / √x.

Using the quotient rule, the derivative is given by:

dy/dx = [(√x)(1) - (x + 4)(1/2x^(-1/2))] / (√x)^2

Simplifying this expression, we get:

dy/dx = (2 - (x + 4)/(2√x)) / x

Next, we need to evaluate the derivative at the point (1,5).

Substituting x = 1 into the derivative expression, we have:

dy/dx = (2 - (1 + 4)/(2√1)) / 1
       = (2 - 5/2) / 1
       = (4 - 5) / 2
       = -1/2

So, the slope of the tangent line at (1,5) is -1/2.

Using the point-slope form of a line, we can write the equation of the tangent line as:

y - 5 = (-1/2)(x - 1)

Simplifying this equation, we get:

y - 5 = -1/2x + 1/2

Rearranging, we have:

y = -1/2x + 11/2

Therefore, the equation of the line tangent to the function y = (x + 4) / √x at the point (1,5) is y = -1/2x + 11/2.

Charles Arocha · Answer

undefined To find the equation of a line tangent to the function $ y = \frac{x + 4}{\sqrt{x}} $ at the point $ (1, 5) $, follow these steps:

1. Find the derivative of the function $ y = \frac{x + 4}{\sqrt{x}} $ using the quotient rule.

2. Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line at the point $ (1, 5) $.

3. Use the point-slope form of the equation of a line, $ y - y_1 = m(x - x_1) $, where $ m $ is the slope found in step 2 and $ (x_1, y_1) $ is the given point $ (1, 5) $, to write the equation of the tangent line.

4. Simplify the equation to obtain the final form.

Let's go through these steps:

1. Derivative of $ y = \frac{x + 4}{\sqrt{x}} $:
$$ y' = \frac{d}{dx}\left(\frac{x + 4}{\sqrt{x}}\right) $$

2. Evaluate the derivative at $ x = 1 $ to find the slope:
$$ y'(1) = \frac{d}{dx}\left(\frac{x + 4}{\sqrt{x}}\right)\bigg|_{x=1} $$

3. Plug $ x = 1 $ into the derivative expression and simplify to find $ y'(1) $, which represents the slope of the tangent line at $ (1, 5) $.

4. Use the point-slope form to write the equation of the tangent line:
$$ y - 5 = m(x - 1) $$

5. Replace $ m $ with the slope found in step 3, and $ x_1 $ and $ y_1 $ with the coordinates of the given point $ (1, 5) $.

6. Simplify the equation to obtain the final form.

This process will yield the equation of the tangent line to the function at the point $ (1, 5) $.