y = x^3 - 8x^2 + 6x -2 has a slope of 6 at two points, what are the coordinates?
I did the power rule to derive the equation: y'= 3x^2 - 8x + 6 , and then I tried to factor it to find the x intercepts but it doesn't work. But anyway, my point of confusion is where does the slope of 6 come in all of this? I know I may need the point slope formula, but can anyone give some tips of how to go about this problem?
You are told that the slope is 6, and you know that the derivative is the slope, so ...
3x^2 - 15x + 6 = 6 , your y' has an error
3x^2 - 15x = 0
3x(x - 5) = 0
x = 0 or x = 5
if x = 0, y = -2
if x = 5, y = 125 - 200 + 30 - 2 = -47
So the two points are (0,-2) and (5,-47)
It did not ask for the equation of those tangents, but the first one is easy to find, since it is a y-intercept, so
y = 6x - 2
I will check my work by plotting this line and the original curve.
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3+-+8x%5E2+%2B+6x+-2,+y+%3D+6x-2
looks line a nice tangent.
Thanks for your help! But wouldn't the derivative be y'=3x^2-16x+6 not 3x^2-15x+6?
Then I'll have x(3x-16x)=0
x=0
3x-16x=0 ~> -15x=0 , x=0
I'm confused even more now...
*** wait. I mean x(3x-16)=0, I'm not sure what's wrong with my eyes. But I will try to figure out how to do it now. Thank you again
Of course you are right, looks like I had a typo and carried it through,
so you would end up with
3x^2 - 16x = 0
x(3x - 16) = 0
x = 0, or x = 16/3
the only change would be ...
when x = 16/3
y = (16/3)^3 - 8(16/3)^2 + 6(16/3)- 2 = -1238/27
so the 2nd point is (16/3, -1238/27)
To find the coordinates where the slope of the given equation y = x^3 - 8x^2 + 6x - 2 equals 6, you will need to find the values of x that satisfy the equation y' = 6.
You correctly derived the equation for the derivative, y' = 3x^2 - 8x + 6. Now, let's set this derivative equal to 6:
3x^2 - 8x + 6 = 6
Simplifying this equation, we get:
3x^2 - 8x = 0
Factoring out an x, we have:
x(3x - 8) = 0
Setting each factor equal to zero, we find two possible values for x:
x = 0
and
3x - 8 = 0 => 3x = 8 => x = 8/3
So, we have two potential x-values: x = 0 and x = 8/3.
To find the corresponding y-coordinates, substitute these x-values back into the original equation:
For x = 0:
y = (0)^3 - 8(0)^2 + 6(0) - 2 = -2
For x = 8/3:
y = (8/3)^3 - 8(8/3)^2 + 6(8/3) - 2
Evaluating this expression, you will find the corresponding y-coordinate.
Therefore, the coordinates where the slope of the equation y = x^3 - 8x^2 + 6x - 2 is equal to 6 are (0, -2) and (8/3, y-coordinate).
Note: Finding the y-coordinate for x = 8/3 might require additional calculations, but this is the general process to find the coordinates.