Find the tangents that meet the curve x^3-5x^2+2x+8 at the gradient = 10
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.
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
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.
To find the tangents that meet the curve at a specific gradient, we need to find the points on the curve where the gradient is equal to the given value.
The gradient of a curve at a point can be found by taking the derivative of the curve equation.
So, let's find the derivative of the curve equation x^3 - 5x^2 + 2x + 8.
Taking the derivative of the curve equation, we get:
f'(x) = 3x^2 - 10x + 2.
Now, we can set f'(x) equal to the given gradient of 10:
3x^2 - 10x + 2 = 10.
Rearranging the equation:
3x^2 - 10x + 2 - 10 = 0,
3x^2 - 10x - 8 = 0.
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = 3, b = -10, and c = -8.
Substituting these values into the quadratic formula, we get:
x = (-(-10) ± √((-10)^2 - 4(3)(-8))) / (2(3)),
x = (10 ± √(100 + 96)) / 6,
x = (10 ± √196) / 6.
Simplifying further:
x = (10 ± 14) / 6,
x = 24/6 or -4/6.
x = 4 or x = -2/3.
Now that we have the x-values, we can substitute them back into the original curve equation to find the corresponding y-values.
For x = 4:
y = 4^3 - 5(4)^2 + 2(4) + 8,
y = 64 - 80 + 8 + 8,
y = 0.
So, one point on the curve with gradient = 10 is (4, 0).
For x = -2/3:
y = (-2/3)^3 - 5(-2/3)^2 + 2(-2/3) + 8,
y = -8/27 - 20/9 - 4/3 + 8,
y = -8/27 - 60/27 - 36/27 + 216/27,
y = 112/27.
So, another point on the curve with gradient = 10 is (-2/3, 112/27).
Therefore, the two tangents that meet the curve x^3 - 5x^2 + 2x + 8 at the gradient of 10 are given by the points (4, 0) and (-2/3, 112/27).