Find an equation of the line that is tangent to the graph of f and parallel to the given line.
Function
f(x) = 2x^2
Line
6x – y + 4 = 0
f'(x) = 4x
slope of given line = 6
so
4x = 6
x = 3/2
then y = 9/2
since new line is parallel to old line, it should look like
6x- y= K
plug in point (3/2, 9/2)
6(3/2) -9/2 =k
k = 9/2
so 6x - y = 9/2, or
12 - 2y = 9
To find an equation of the line that is tangent to the graph of f and parallel to the given line, we need to use the fact that parallel lines have the same slope.
First, find the slope of the given line. The given line is in the standard form Ax + By + C = 0, where A, B, and C are constants. The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept.
To convert the given line into slope-intercept form, isolate y:
6x - y + 4 = 0
-y = -6x - 4
y = 6x + 4
From this equation, we can see that the slope of the given line is 6.
Since we want the tangent line to be parallel to the given line, the tangent line will also have a slope of 6.
Next, find the derivative of the function f(x) = 2x^2. The derivative gives us the slope of the tangent line at any point on the graph.
f'(x) = d/dx(2x^2)
f'(x) = 4x
Now, set the derivative equal to the slope we found earlier (6) and solve for x.
4x = 6
x = 6/4
x = 3/2
We now have the x-coordinate of the point where the tangent line touches the graph of f. To find the y-coordinate, substitute this x-value back into the original function f(x):
f(3/2) = 2(3/2)^2
f(3/2) = 2(9/4)
f(3/2) = 9/2
Therefore, the point where the tangent line touches the graph of f is (3/2, 9/2).
Now, we have the slope (6) and a point (3/2, 9/2) that lies on the tangent line. We can use the point-slope formula to find the equation of the line:
y - y1 = m(x - x1)
Using the values we have:
y - 9/2 = 6(x - 3/2)
Expanding the equation:
y - 9/2 = 6x - 9
Simplifying and rearranging:
y = 6x - 9 + 9/2
y = 6x - 9/2
Therefore, the equation of the line tangent to the graph of f and parallel to the given line is y = 6x - 9/2.