What are the coordinates of the point on the graph of f(x)=(x+1)(x+2) at which the tangent is parallel to the line with equation 3x-y-1=0?
a)(-3,2)
b)(-1,0)
c)(0,2)
d)(1,6)
To answer this, I put f(x)=(x+1)(x+2) into the form f(x)=x^2+3x+2 and found the derivative which is f'(x)=2x+3. Next I re-arranged the equation 3x-y-1=0 and got y=3x-1. I don't know what to do after that
I found the derivative of y=3x-1 which is y=3. Then I had the two differentiated equations equal one another:
2x+3=3 and solve for x: x=0
Then I subbed x into f(x)=x^2+3x+2 and got y=2
So the coordinate is (0,2)?
good work. Confirm with the graphs at
http://www.wolframalpha.com/input/?i=plot+x%5E2%2B3x%2B2%2C+y%3D3x%2B2+for+-1+%3C%3D+x+%3C%3D+1
To find the point on the graph of f(x) = (x + 1)(x + 2) at which the tangent is parallel to the line with equation 3x - y - 1 = 0, we can follow these steps:
1. Convert the equation of the line 3x - y - 1 = 0 into slope-intercept form (y = mx + b), where m represents the slope and b represents the y-intercept:
Rearranging the equation, we get: y = 3x - 1.
Therefore, the slope of the line is 3.
2. Find the derivative of f(x) by differentiating f(x) = (x + 1)(x + 2) with respect to x. The derivative will give us the slope of the tangent line at any given point on the graph of f(x):
Differentiating f(x) = (x + 1)(x + 2), we get: f'(x) = 2x + 3.
3. Set the slope of the tangent line (given by the derivative) equal to the slope of the given line, and solve for x:
Since the tangent line is parallel to the given line, the slopes will be equal. Thus, we have the equation:
2x + 3 = 3
Solving for x, we get: x = 0.
4. Plug the value of x into the equation of f(x) to find the corresponding y-coordinate:
Evaluating f(x) = (x + 1)(x + 2) with x = 0, we get: f(0) = (0 + 1)(0 + 2) = 2.
Therefore, the point on the graph of f(x) at which the tangent is parallel to the line 3x - y - 1 = 0 is (0, 2), which corresponds to option c) (0, 2).