+31 Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1 and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2). Find an equation of the line tangent to the following curves at x = 2.
a. y = f(x) + g(x)
b. y = f(x) - 2g(x)
c. y = 4f(x)
I don't even know where to start with this one. Thanks for helping!
+118609 Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1
This tells you that the point (2,9) is on the f graph
\(f(2)=9\)
The Equation of tangent to the f funtion at x=2 is y=4x+1
and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2).
At f'(2) = 3
And this tangent line, the one at f(2) passes through (0,-2)
For this line we are told (0,-2) is a point and the gradient is 3. We need to find the find the y value for x=2.
\(m=\frac{rise}{run}\\ 3=\frac{y_1--2}{2-0}\\ 6=y_1+2\\ y_1=4\\ g(2)=4 \)
The Equation of tangent to the g funtion at x=2 is y=3x-2
a. y = f(x) + g(x)
f'(2)+g'(2)=4+3=7
f(2)+g(2)= 9+4=13
So the equation of the tangent at x=2 is
y=7x+13
b. y = f(x) - 2g(x)
2g(x) the gradient and the y values will be doubled so gradient of tangent=6 at (2,8)
f'(2)+2g'(2)=4+6=10
f(2)+g(2)= 9+8 = 17
y=10x-17
c. y = 4f(x)
tangent of 4f(x)=4(4x+1) at x=2 is y = 16x+4
I think that is right.
You can try convincing yourself (one way or the other) by drawing some graphs.
