Find an equation of the line that is tangent to the graph of f and parallel to the given line
Function: f(x) = x^3
Line 3x-y+1 = 0
._.; I know the derivative of the function is 3x^2 and I do have the answer in the back of my book just need the way to get it..
Sorry in advance-- I don't know how to find the answer, but I'm willing to help all I can.
First, rearrange the equation of the line to make it y=-3x-1. Since we need to find a line parallel to this one, we know it needs a slope of -3.
y=-3x-1
y=mx+b
m=slope
m=-3
What is the answer the back of the book gives? Sometimes it's possible to work backwards from the answer.
Sorry that's as far as I could go.
I'll try to help you out all I can--I'm trying to work it out right now.
--Cori
:3 yep I got there too.
It was y=3x+2 I think.
It doesn't matter though XD I'll just ask my teacher when I get there
To find an equation of the tangent line to the graph of the function f(x) = x^3 that is parallel to the given line 3x-y+1 = 0, we need to follow these steps:
1. Find the slope of the given line: rearrange the equation 3x - y + 1 = 0 to isolate y: y = 3x + 1. Notice that the slope of this line is 3 since the equation is in the form y = mx + b, where m represents the slope.
2. Determine the derivative of the function f(x) = x^3: taking the derivative of x^3 with respect to x gives us 3x^2. The derivative represents the slope of the tangent line at any point on the graph of f(x).
3. Since we want the tangent line to be parallel to the given line, the slope of the tangent line must be equal to 3.
4. Now we have the slope of the tangent line (m = 3) and a point on the graph of f(x) (x, f(x)). To find the point (x, f(x)), we need to substitute x into the original function, f(x) = x^3. Let's assume x = a for now.
5. So, the coordinates of the point on the graph are (a, f(a)) = (a, a^3).
6. Use the point-slope form of the equation of a line to get the equation of the tangent line: y - y1 = m(x - x1). Substituting in the values we have, we get: y - a^3 = 3(x - a).
7. Simplify the equation to obtain the final equation of the tangent line: y - a^3 = 3x - 3a.
Therefore, the equation of the tangent line to the graph of f(x) = x^3 that is parallel to the line 3x - y + 1 = 0 is y = 3x - 3a + a^3, where a is the x-coordinate of the point on the graph.