Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function Line
f(x) = 2x^2 6x – y + 4 = 0

I know i have to find derivative 1st which will be f'(x)=4x but then how to i find equation do i plug 6 in for x so 4(6) would it be 6x-24???

To find the equation of a line that is tangent to the graph of a function and parallel to a given line, you need two pieces of information: the slope of the tangent line and a point that lies on the tangent line.

1. Start by finding the derivative of the function f(x) = 2x^2 to determine the slope of the tangent line. The derivative of f(x) = 2x^2 is f'(x) = 4x.

2. Since the tangent line is parallel to the given line, they must have the same slope. The given line has a slope of 6.

3. Set the derivative of f(x) equal to the slope of the tangent line to find the x-value where the tangent line intersects the graph of f(x).

4x = 6
x = 6/4
x = 3/2

4. Plug the x-value into the original function f(x) = 2x^2 to find the y-value.

f(3/2) = 2(3/2)^2
= 2(9/4)
= 18/4
= 9/2

So, the point (3/2, 9/2) lies on the tangent line.

5. Use the point-slope form of a linear equation to write the equation of the tangent line:

y - y1 = m(x - x1)

where m is the slope of the tangent line and (x1, y1) is the point on the tangent line.

Plugging in the values, we have:

y - (9/2) = 6(x - (3/2))

6. Simplify the equation:

y - 9/2 = 6x - 9
y = 6x - 9 + 9/2
y = 6x - 9/2

Therefore, the equation of the line that is tangent to the graph of f and parallel to the given line 6x – y + 4 = 0 is y = 6x - 9/2.