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

y = 3x - 2 (smaller y-intercept)

y = 3x + 2 (larger y-intercept)

Step-by-step explanation:

First let's write the generic equation of a line:

y = ax + b

This line needs to be parallel to the line 3x - y + 5 = 0, so it needs to have the same slope of this line.

The line 3x - y + 5 = 0 has a slope of 3, so our line has a = 3:

y = 3x + b

Now we need to find the values of b that make this line tangent to the function f(x) = x^3

Let's first find the derivative of f(x) in relation to x:

df(x)/dx = 3x^2

This derivative is the slope of the tangent line to the function for any value of x. We need a slope of 3, so:

3x^2 = 3

x^2 = 1

x = ±1

Now, to find the y-values, we have:

f(1) = 1^3 = 1

f(-1) = (-1)^3 = -1

So, using the points (1,1) and (-1,-1) in our parallel line, we have:

first line using (1,1) : 1 = 3*1 + b

b = -2

second line using (-1,-1) : -1 = -3*1 + b

b = 2

The value of b is the y-intercept of the line, so the line with smaller y-intercept is y = 3x - 2, and the line with larger y-intercept is y = 3x + 2

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

im alright what do you need

step-by-step explanation: