2015年2月11日 · The point at which the tangent line is horizontal is (-2, -12). To find the points at which the tangent line is horizontal, we have to find where the slope of the function is 0 because a horizontal line's slope is 0. d/dxy = d/dx(16x^-1 - x^2) d/dxy = -16x^-2 - 2x That's your derivative. Now set it equal to 0 and solve for x to find the x values at which the tangent line is horizontal …

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. So if the function is f(x) and if the tangent "touches" its curve at x=c, then the tangent will pass through the point (c,f(c)). The slope of this tangent line is …

2014年7月23日 · The slope of a horizontal tangent line is 0. To calculate the slope of a straight line, we take a difference in the y dimension and divide it by the change in the x dimension of two points on the line: "slope" = (y_1 - y_2)/(x_1 - x_2) assuming points (x_1, y_1) and (x_2, y_2) lie on the line For a horizontal line y_1 - y_2 = 0 so "slope" = 0/(x_1 - x_2) = 0

2018年3月18日 · How do you find the Tangent line to a curve by implicit differentiation? What is the slope of a line tangent to the curve #3y^2-2x^2=1#? How does tangent slope relate to the slope of a line?

2016年12月16日 · A horizontal tangent will have a 0 slope. Therefore, we set the derivative to 0 and solve. 0 = 1 + 2cosx -1/2 = cosx x = (2pi)/3 and (4pi)/3 Hopefully this helps! Trigonometry

2016年10月23日 · To solve this question, we can sub in -2x wherever we see a y in our original equation (substitution method). x^2 + x(-2x) + (-2x)^2 = 1 x^2 -2x^2 +4x^2 = 1 3x^2 = 1 x^2 = 1/3 x = +- sqrt(1/3) STEP 5: Now that we know the x-value of the point, we can easily find the y-value of the point because we know y=-2x where the tangent line is horizontal.

#x^2y-2x+18y=3# Enter your answers as a comma separated list of ordered pairs (x,y) Calculus. 1 Answer

Find all values of x for which the following function has a horizontal tangent line. y = x^3 e^x; Find all point(s) on 2yx^2 + y^3 = 3 that have a horizontal tangent line. Find all points where the tangent line to the following curve is horizontal: r = 1 - sin \theta.

Find the points on the curve r = 1 + \cos \theta where the tangent line is horizontal or vertical. Find the points on the curve r = 8 \cos \theta where the tangent line is horizontal or vertical. Find the points of the curve where the tangent line is horizontal or vertical. x = t^3 - 12 t; y = \frac{t^3}{3} - \frac{t^2}{2} + 1

Horizontal Tangent Line. A horizontal tangent line is a straight line that only touches the function at one point, this straight line can be vertical, horizontal or a slant straight line, also, the slope of the tangent line is given by the first derivative. Answer and Explanation: 1