How do you find the equation of the line tangent to the graph of #y = sqrtx# at (16,4)?
By signing up, you agree to our Terms of Service and Privacy Policy
To find the equation of the line tangent to the graph of y = sqrt(x) at the point (16,4), we can use the derivative of the function.
First, we find the derivative of y = sqrt(x) with respect to x, which is 1/(2*sqrt(x)).
Next, we substitute x = 16 into the derivative to find the slope of the tangent line at that point.
The slope of the tangent line is 1/(2*sqrt(16)) = 1/8.
Using the point-slope form of a linear equation, we have y - 4 = (1/8)(x - 16).
Simplifying the equation, we get y = (1/8)x + 2.
Therefore, the equation of the line tangent to the graph of y = sqrt(x) at (16,4) is y = (1/8)x + 2.
By signing up, you agree to our Terms of Service and Privacy Policy
To find the equation of the line tangent to the graph of (y = \sqrt{x}) at the point (16,4), you need to find the slope of the tangent line at that point, then use point-slope form to write the equation of the tangent line.
-
Find the derivative of the function (y = \sqrt{x}). [ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} ]
-
Evaluate the derivative at x = 16 to find the slope of the tangent line at that point. [ \frac{dy}{dx} \bigg|_{x=16} = \frac{1}{2\sqrt{16}} = \frac{1}{8} ]
-
Use the point-slope form of a line with point (16,4) and slope (\frac{1}{8}). [ y - y_1 = m(x - x_1) ] [ y - 4 = \frac{1}{8}(x - 16) ]
-
Simplify to get the equation in slope-intercept form, if necessary. [ y - 4 = \frac{1}{8}x - 2 ] [ y = \frac{1}{8}x + 2 ]
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- What is the slope of the line normal to the tangent line of #f(x) = secx-cos^2(x-pi) # at # x= (pi)/12 #?
- How to find instantaneous rate of change for # f(x) = x³# when x=2?
- What is the equation of the line normal to #f(x)=(x-1)^2-2x+5# at #x=-2#?
- What is the slope of the line normal to the tangent line of #f(x) = tanx+sin(x-pi/4) # at # x= (5pi)/6 #?
- Using the limit definition, how do you differentiate #f(x)=4 -2x -x^2#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7