find the gradient of the curve y=2x^3-5x^2-x+1 at the pt (2,-5). find the x-coordinate of another pt on the curve where the tangent at that pt is parallel to the tangent at the pt (2,-5)
rather long question.....
dy/dx = 6x^2 - 10x - 1
at (2,-5) , dy/dx = 3
equation of tangent at that point:
y = 3x+b, with (2,-5) on it
-5 = 6+b
b = -11
y = 3x - 11
so now intersect: set y = y
2x^3 - 5x^2 - x + 1 = 3x - 11
2x^3 - 5x^2 -4x + 12 = 0
We already know that x=2 is a solution
so x-2 is a factor.
by synthetic division I found
2x^3 - 5x^2 -4x + 12 = (x-2)(2x^2 - x - 6)
= (x-2)(x-2)(2x+3)
we have a double root, which is the case where a tangent touches the curve, and another at
x = -3/2
y = ...... You do the arithmetic.
To find the gradient of the curve at the point (2, -5), we will find the derivative of the curve and substitute x = 2 into it.
1. First, differentiate the equation y = 2x^3 - 5x^2 - x + 1 with respect to x. The power rule for differentiation states that if y = ax^n, then dy/dx = anx^(n-1). Applying this rule:
dy/dx = d/dx (2x^3) - d/dx(5x^2) - d/dx(x) + d/dx(1)
= 6x^2 - 10x - 1 + 0
= 6x^2 - 10x - 1
2. Substituting x = 2 into the derivative equation, we can find the slope (gradient) at the point (2, -5).
dy/dx = 6(2)^2 - 10(2) - 1
= 24 - 20 - 1
= 3
So, the gradient of the curve y = 2x^3 - 5x^2 - x + 1 at the point (2, -5) is 3.
To find the x-coordinate of another point on the curve where the tangent is parallel to the tangent at (2, -5), we need to find the equation of the tangent line at (2, -5), and then find another point on the curve with the same slope.
3. The equation of a tangent line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the curve and m is the slope.
For our given point (2, -5), the equation of the tangent line becomes y + 5 = 3(x - 2).
4. Simplifying the equation, we get y = 3x - 11.
Since the slope of the tangent line is 3, we need to find another point on the curve with a slope of 3.
5. To find this point, we'll set dy/dx = 3 and solve the resulting equation.
6x^2 - 10x - 1 = 3
6x^2 - 10x - 4 = 0
Factoring out 2:
2(3x^2 - 5x - 2) = 0
Applying the quadratic formula:
x = (-(-5) ± √((-5)^2 - 4(3)(-2))) / (2(3))
x = (5 ± √(25 + 24)) / 6
x = (5 ± √49) / 6
x = (5 ± 7) / 6
This gives us two potential x-values: (5 + 7)/6 and (5 - 7)/6.
x = 12/6 and x = -2/6
Simplifying, we get:
x = 2 and x = -1/3
Therefore, there are two potential x-coordinates for points where the tangent is parallel to the tangent at (2, -5): x = 2 and x = -1/3.