Q&A for How to Find the Equation of a Tangent Line

Take your first derivative, then plug the x-value into the point you want a tangent line.

The equation for a line is, in general, y=mx+c. To find the equations for lines, you need to find m and c. m is the slope. For example, if your line goes up two units in the y direction, for every three units across in the x direction, then m=2/3. If you have the slope, m, then all you need now is c. To find c in any line, you can use any (x,y) points you know. In the case of a line that is tangent to a graph, you can use the point (x,y) where the line touches the graph. If you use that x and that y and the slope m, you can use algebra to find c. y=mx+c, so, c=y-mx. Once you have c, you have the equation of the line! Done.

Parallel lines always have the same slope, so since y = 2x + 3 has a slope of 2 (since it's in slope-intercept form), the tangent also has a slope of 2. Now you also know that f'(x) will equal 2 at the point the tangent line passes through. Differentiate to get the equation for f'(x), then set it equal to 2. Now you can solve for x to find your x-coordinate, plug that into f(x) to find the y-coordinate, and use all the information you've found to write the tangent line equation in point-slope form.

Unless you are given the slope of the tangent line, you'll need to find it the same way you would for any other problem: finding the derivative f'(x). Trigonometric functions have their own rules for differentiation, which you can look up in your textbook or online. To get you started, the derivative of sin(x) is cos(x).

Two lines are perpendicular to each other if the product of their slopes is -1. So to find the equation of a line that is perpendicular to the tangent line, first find the slope of the tangent line. Let's call that t. If the slope of the line perpendicular to that is p, then t*p=-1, or p=-1/t. To get the whole equation of the perpendicular, you need to find a point that lies on that line, call it (x°, y°). You already know the slope p and have a point (x°, y°), so you fill those in the standard equation of a line y=mx+b to get y°=px°+b. Solve this for b. Then the final equation of the line perpendicular to your tangent line is y=px+b.

If you're referring to the graph of the equation x = 1, that's a straight line and does not have a tangent. (Only curved lines have tangents.)

If you know the point-slope form of a line, you can write the answer down directly since you are given a point (-8, -4) on and the slope (-10) of the line. That would be: (y - (-4)) = -10 (x - (-8)). If you prefer it in a different form, such as slope-intercept, you can convert into that form. Alternatively, you could try to calculate the y-intercept directly by taking the point (-8, -4) as a starting point. If you move 8 units right to the y-axis, then you move 80 units down to (0, -84), so y = -10x - 84.

Differentiate the equation of the circle and plug in the values of x,y in the derivative. This gives you the slope of the tangent at (x,y). Use the slope-point form of the line to find the equation, with the slope you obtained earlier and the coordinates of the point.

Differentiate the curve at the point given to obtain the slope, then substitute the slope and the point in the slope-point form to get the equation.