Let f(x)=(x^2+1)*(2x-3)
Find the equation of the line tangent to the graph of f(x) at
x=3.
Find the value(s) of x where the tangent line is horizontal.
If f(x)=2x^2−5x+3, find
f'(−4).
Use this to find the equation of the tangent line to the parabola
y=2x^2−5x+3 at the point (−4,55). The equation of this tangent line
can be written in the form y=mx+b
where m is: ????
and where b is: ????
1. Find the equation of the line tangent to the curve y=2x^2 +
sin4x at x= π/3.
2. Determine the point(s) where the tangent line to y= 2sinx-4x
has a slope of-3 in the domain 0≤x≤ 2π.
b) Find the equation of the tangent line.
Part A: Find the slope of the tangent line to the graph of
f(x)=√(5x+9) at the point (8,7)
Part B: Find an equation of the tangent line to the graph of
f(x)= −3x^2 at the point (−3,−27). Solve your equation for ?.
Part C: Find an equation of the tangent line to the graph of
?(?)= 7/(x-4) at the point (5,7). Solve your equation for ?.