How do you find the equation of the tangent line to the graph of #f(x)=x^2+1# at point (2,5)?

Question

Emma Aldridge · Accepted Answer

#y=4x-3# #m_(color(red)"tangent")=f'(x)" at x = 2"# #rArrf'(x)=2xrArrf'(2)=4# #y-5=4(x-2)# #rArry=4x-3larrcolor(blue)"is equation of tangent"#

Aurora Alvarado · Answer

undefined To find the equation of the tangent line to the graph of f(x)=x^2+1 at point (2,5), 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.

Taking the derivative of f(x)=x^2+1, we get f'(x) = 2x.

Substituting x=2 into f'(x), we find f'(2) = 2(2) = 4.

Therefore, the slope of the tangent line at point (2,5) is 4.

Using 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, we can substitute the values to find the equation of the tangent line.

Plugging in the values (2,5) and m=4, we have y - 5 = 4(x - 2).

Simplifying the equation, we get y - 5 = 4x - 8.

Rearranging the equation, we have y = 4x - 3.

Therefore, the equation of the tangent line to the graph of f(x)=x^2+1 at point (2,5) is y = 4x - 3.