How do you find the equation of the tangent line to a given equation fx equals x3 plus 3x2-5 that is perpendicular to a given equation 2x-6y plus 1 equals 0 with no x values given?

Answer 1

First consider the given line: 2x - 6y + 1 = 0

Subtract (2x+1) from each side: -6y = -2x - 1

Divide each side by -6: y= 1/3 x + 1/6

The slope of this line is 1/3, so any line perpendicular to it has a slope of -3 .

Now let's find places on the function where its slope is -3 :

f(x) = x3 + 3x2 - 5

The slope of the tangent to the function at every point is the value of

its first derivative at that point.

f'(x) = 3x2 + 6x and we want it to be = -3 .

3x2 + 6x + 3 = 0

Divide both sides by 3 to make it neater and easier:

x2 + 2x + 1 = 0

Either use the quadratic equation to solve this, or else notice that

it's a perfect square ... (x+1)2 = 0, so

x = 1

That's the x-value on the graph of the function where the slope of the

tangent to the function is -3.

Substitute this 'x' back into the original function to find the corresponding 'y' value.

Then you'll have the 'x' and 'y' coordinates of the point on the graph of the function

where the tangent has the slope of -3 .

Now you have the slope and one point on the line that solves the problem,

and finding the equation of the line should now be a piece o' cake.