Write the equation for the tangent line to the graph of f(x)=x^2-3 where x=2

Answer 1

f(x) = x^2 -3

take the derivative = slope

f'(x) = 2x

when x=2

f'(2) = 2(2) = 4 = the slope of the tangent line at the point (2, f(2)) = (2, 2^2-3) = (2,4-3) = (2,1)

find the line with slope =4, through the point (2,1)

y-1 = 4(x-2) in point slope form

or

y= 4x -7 in slope intercept form

it helps to sketch a graph. It's an upward opening parabola, with y intercept = minimum point = vertex (0,-3) and goes through two other points (2,1) and (-2,1) with axis of symmetry x=0. Tangent line at (2,1) has slope=4 and y intercept (0,-7)