one more calculus question?

Question

i need to find the equation of the line that is tangent to the given function and parallel to the given line. please outline the steps with explinations, thanks.

f (x) = (x^3) + 2

3x - y - 4 = 0 or in other words y = 3x - 4

Answer 1

If you remember from algebra parallel lines have equal slopes.
The slope of your tangent line is 3
f'(x) = 3x²

So you need to have a point to find the rest of the tangent line.
So, 3x² = 3 or x = ±1, your question asked for only one tangent line, so I will do the part for x = 1
Evaluate f(x) at x = 1,
f(x) = x^3 + 2 = 1 + 2 = 3
So the point that you need is (1,3).
The slope of your tangent line is 3:
The equation is:
y = 3x + b
You need to find the y-intercept, plug in the point (1,3)
3 = 3 + b
b = 0
The equation of the tangent line is y = 3x

Answer 2

Alright, so you know the tangent (a line) must be parallel to a given line. So, that means you know the slope of tangent.

Now, you need to figure out at what point that function has that slope. For example, if the equation was x^5+x and you needed a slope of 2, you'd take the derivative=5x^4+1. Now solve, 2=5x^4+1. So x, in this case = (2/5)^(1/4). Wow, what would be really ugly.

So now, you need to find the y coordinant that corresponds to that x coordinant. In the example above, you'd plug (2/5)^(1/4) back into the x^5+x equation.

So what I've done is first found the slope. Now I've found the point on the graph that has that slope. Use whatever algebra 1 you want to make a line with a certain slope that goes through a particular point.

Hope this helps!
Catherine

Answer 3

slope of a tangent f´(x) = 3x^2 = 3
so x=+-1 so you get the points
(1,3) and (-1,3)
y-3=3(x-1) and y-3=3(x+1)