How do you find the equation of a line tangent to a graph #f(x)=2x^2# at (3,1)?

Question

Aurora Alvarado · Accepted Answer

In the pint slope form it would be y-1=12(x-3)

Find the slope first, f'(x) = 4x. At (3,1) slope would be 4(3)=12. Equation of the line tangent at (3,1) in the point slope form would be y-1=12(x-3)

Christopher Allers · Answer

undefined To find the equation of a line tangent to a graph at a specific point, you need to determine the slope of the tangent line at that point. To do this, you can find the derivative of the function and evaluate it at the given x-coordinate.

The derivative of f(x) = 2x^2 is f'(x) = 4x.

To find the slope of the tangent line at x = 3, substitute x = 3 into the derivative: f'(3) = 4(3) = 12.

The slope of the tangent line is 12.

Next, use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values (3, 1) and m = 12 into the equation, we get: y - 1 = 12(x - 3).

Simplifying the equation gives the equation of the line tangent to the graph f(x) = 2x^2 at (3, 1): y = 12x - 35.