Question

Find the tangents that meet the curve x^3-5x^2+2x+8 at the gradient = 10

Answer 1

AAAaannndd the bot gets it wrong yet again!

y = -14x + 68
has a slope of -14, so it cannot be tangent with a gradient of 10.
f'(4) = 10, so the tangent line is
y-f(4) = 10(x-4)
y = 10x-40

and f(-2/3) = 112/27, so the tangent there is
y - 112/27 = 10(x + 2/3)
y = 10x + 292/27

Answer 2

To find the tangents that meet the curve at a gradient of 10, we need to first find the derivative of the curve.

f(x) = x^3 - 5x^2 + 2x + 8

f'(x) = 3x^2 - 10x + 2

We can now set f'(x) to 10 and solve for x to find the point(s) on the curve where the gradient is 10.

3x^2 - 10x + 2 = 10
3x^2 - 10x - 8 = 0
(3x + 2)(x - 4) = 0

So x = -2/3 or x = 4.

Now we need to find the equation(s) of the tangent(s) at these points.

At x = -2/3:

The gradient of the curve at this point is:

f'(-2/3) = 3(-2/3)^2 - 10(-2/3) + 2 = 10

So the tangent at this point has gradient of 10.

Using the point-gradient equation of a line:

y - f(-2/3) = 10(x + 2/3)

We can simplify this equation by first finding f(-2/3):

f(-2/3) = (-2/3)^3 - 5(-2/3)^2 + 2(-2/3) + 8 = 184/27

So the equation of the tangent at x = -2/3 is:

y - 184/27 = 10(x + 2/3)

y = 10x + 566/27

At x = 4:

The gradient of the curve at this point is:

f'(4) = 3(4)^2 - 10(4) + 2 = -14

So the tangent at this point has gradient of -14.

Using the point-gradient equation of a line:

y - f(4) = -14(x - 4)

We can simplify this equation by first finding f(4):

f(4) = 4^3 - 5(4)^2 + 2(4) + 8 = -12

So the equation of the tangent at x = 4 is:

y + 12 = -14(x - 4)

y = -14x + 68

Therefore, the two tangents that meet the curve at a gradient of 10 are:

y = 10x + 566/27 and y = -14x + 68.

Answer 3

I apologize for the mistake in my previous response. You are correct that the equation y = -14x + 68 cannot be the tangent line at x = 4 since its slope is -14, not 10. Thank you for pointing out the error.

The correct tangent line at x = 4 is indeed y = 10x - 40, since f'(4) = 10, as you stated.

Similarly, the correct tangent line at x = -2/3 is y = 10x + 292/27, with f'(-2/3) = 10.

Thank you for bringing this to my attention and allowing me to correct my mistake.