Question

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Where k is a constant.
Given that passes through the points (0, -2) and (2, 18).
a. Show that k = 2 and find an equation for C
b. Show that the line with equation y = x-2 is a tangent to C and find the coordinates of the point of contact.​

Answer 1

The constant k is found to be 2 by integrating the derivative and using the given points that the curve passes through. The equation of the curve C is y = x³ + 2x² + 2x - 2. To find the point of contact for the tangent line y = x - 2, equate its slope to the curve's derivative and solve for x and y.

Step-by-step explanation:

To find the constant k and an equation for the curve C, we start by integrating the given derivative dy/dx = 3x² + 4x + k. When we integrate, we obtain the function y = x³ + 2x² + kx + c, where c is the integration constant. Since the curve passes through the point (0, -2), substituting x = 0 and y = -2 gives us c = -2. When the curve passes through the point (2, 18), we can solve for k by substituting x = 2 and y = 18 into our equation, which yields k = 2. Thus, the equation of the curve C is y = x³ + 2x² + 2x - 2.

For the line with the equation y = x - 2 to be a tangent to the curve C, its slope must be equal to the derivative of C at the point of contact. Substituting the slope (1) of the tangent line into the derivative dy/dx, we find that 3x² + 4x + 2 = 1 yields a solution for x. Solving this equation gives the x-coordinate of the point of contact. Substituting this value back into the equation of the tangent line gives us the y-coordinate.

Answer 2

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

`\displaystyle (dy)/(dx) = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt`

Evaluate the integral to solve for y :

`\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt`

`\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x`

`\displaystyle y = x^3+2x^2+kx - 2`

Use the other known value, f(2) = 18, to solve for k :

`18 = 2^3 + 2*2^2+2k - 2 \implies \boxed{k = 2}`

Then the curve C has equation

`\boxed{y = x^3 + 2x^2 + 2x - 2}`

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

`(dy)/(dx)\bigg|_(x=a) = 3a^2 + 4a + 2`

The slope of the given tangent line
`y=x-2` is 1. Solve for a :

`3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\frac13 \text{ or }a = -1`

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

`x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1`

So, the point of contact between the tangent line and C is (-1, -3).