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How do you find the equation of the line tangent to #y=x^2# at (2,4)?

Answer 1

Emma Aldridge

#y=4x-4#

#•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"#

#y=x^2rArrdy/dx=2x#

#x=2tody/dx=4#

#"using the point-slope form for equation of line"#

#y-4=4(x-2)#

#rArry=4x-4#

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Answer 2

Axel Alvarez

To find the equation of the line tangent to y=x^2 at (2,4), we need to find the slope of the tangent line at that point. The slope of the tangent line is equal to the derivative of the function at that point.

The derivative of y=x^2 is given by dy/dx = 2x.

Substituting x=2 into the derivative, we get dy/dx = 2(2) = 4.

So, the slope of the tangent line at (2,4) is 4.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point (2,4) and m is the slope (4), we can substitute the values to find the equation of the tangent line.

Therefore, the equation of the line tangent to y=x^2 at (2,4) is y - 4 = 4(x - 2).

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Answer from HIX Tutor

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.