What is the slope of the tangent to #y=sqrt(1-x)# at #(-8, 3)#?

Question

Aurora Alvarado · Accepted Answer

#m = -1/6#

Given #y=sqrt(1-x)#

Using the chain rule:

#dy/dx = dy/(dg)(dg)/dx#

where #y = sqrtg# and #g=1-x#, then #dy/(dg) = 1/(2sqrtg) = 1/(2sqrt(1-x))# and #(dg)/dx = -1#

Substituting into the chain rule:

#dy/dx = 1/(2sqrt(1-x)) -1#

#dy/dx = -1/(2sqrt(1-x))#

The slope, #m#, is the derivative evaluated at #x = -8#

#m = -1/(2sqrt(1--8))#

#m = -1/(2sqrt9)#

#m = -1/6#

Samantha Adkison · Answer

undefined To find the slope of the tangent to the curve $ y = \sqrt{1 - x} $ at the point (-8, 3), we first need to find the derivative of the function with respect to $ x $. Then, we evaluate this derivative at the given point to find the slope of the tangent.

1. Find the derivative of $ y = \sqrt{1 - x} $ with respect to $ x $.
2. Evaluate the derivative at the point (-8, 3) to find the slope of the tangent.

The derivative of $ y = \sqrt{1 - x} $ with respect to $ x $ can be found using the power rule and the chain rule:

$$ \frac{dy}{dx} = \frac{d}{dx} \sqrt{1 - x} = -\frac{1}{2\sqrt{1 - x}} \cdot (-1) = \frac{1}{2\sqrt{1 - x}} $$

Now, we evaluate this derivative at the point (-8, 3):

$$ \frac{dy}{dx} \Bigg|_{(-8, 3)} = \frac{1}{2\sqrt{1 - (-8)}} = \frac{1}{2\sqrt{1 + 8}} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \cdot 3} = \frac{1}{6} $$

Therefore, the slope of the tangent to the curve $ y = \sqrt{1 - x} $ at the point (-8, 3) is $ \frac{1}{6} $.