Find an equation of the tangent line at the point specified.
y = $a·sin(x) + $c·cos(x), x = 0
Recall that a tangent line to a graph of y = f(x) at a point P(a, f(a)) is the line through the point P of slope NsDvRziYBBEox5msqhXYNvE2PJWil8aGpRgA1g==. The equation of the tangent line in point-slope form is
y -wPah0ewYTHV1gRKzC8auX7tZI0q0wCSsWA8Elg== = NsDvRziYBBEox5msqhXYNvE2PJWil8aGpRgA1g==(x-a).
We find the formula for the slope of the tangent line to the graph of y = f(x) for any value of x by calculating f'(x).
f(x) = $a·sin(x) + $c·cos(x)
= nc1ItEz0kR4=· + SFgqQUkJGdg= ·
= $a · +oE3dexu+YXOG8QKjTROFi4zaMs= - $c · 6LVZ3Izit10N+67n+zs4eU+LiGc=
(Note the coefficient on the second answer box is -$c, not $c.)
The slope of the tangent line to the graph is given by f'(x) = $a·cos(x) − $c·sin(x). The slope of this line at x = 0 is found by evaluating f'(0).
f'(x) = $a·cos(x) - $c·sin(x)
f'(0) = nc1ItEz0kR4=
Aside from the slope of the tangent line at x = 0, we also need the point that the line passes through on the graph of f(x). Since this line is tangent to f(x) at x = 0, the point needed is (0, f(0)).
Find f(0).
f(x) = $a·sin(x) + $c·cos(x)
f(0) = SFgqQUkJGdg=
Find the equation of the tangent line to f(x) of slope $a at the point (0, $c).
y - f(0) = f'(0)(x-0)
y - SFgqQUkJGdg= = nc1ItEz0kR4= · (x-0)
y = nc1ItEz0kR4= ·x + SFgqQUkJGdg=