Question

10. A curve has the equation y = ax^3 + x + c, where a

Answer 1

Answer:

Part I)

$a=-1\text{ and } c=2$

Giving the equation:

$y=-x^3+x+2$

Part II)

(-1, 2)

Step-by-step explanation:

We have the equation:

$y=ax^3+x+c$

Where a and c are constants.

We are given gradient/slope of the curve at the point (1, 2) is -2.

Part I)

We know that the gradient of the curve exactly at the point (1, 2) is -2.

So, we will differentiate our equation. We have:

$\displaystyle y=ax^3+x+c$

Differentiate both sides with respect to x:

$\displaystyle y^\prime=\frac{d}{dx}\big[ax^3+x+c\big]$

Expand the right:

$\displaystyle y^\prime=\frac{d}{dx}\big[ax^3]+\frac{d}{dx}\big[x\big]+\frac{d}{dx}\big[c\big]$

Since a and c are simply constants, we can move them outside:

$\displaystyle y^\prime=a\frac{d}{dx}\big[x^3]+\frac{d}{dx}\big[x\big]+c\frac{d}{dx}\big[1\big]$

Differentiate. Note that the derivative of a constant is simply 0.

$\displaystyle y^\prime=3ax^2+1$

We know that at the point (1, 2), the gradient/slope is -2.

Hence, when x is 1 and when y is 2, y’ is -2.

So (we don’t have a y so we can ignore it):

$-2=3a(1)^2+1$

Solve for a. Simplify:

$-2=3a+1$

Subtract 1 from both sides yield:

$-3=3a$

So, it follows that:

$a=-1$

Therefore, our derivative will be:

$y=-3x^2+1$

Now, we can go back to our original equation:

$y=ax^3+x+c$

Since we know that a is -1:

$y=-x^3+x+c$

Recall that we know that a point on our curve is (1, 2).

So, when x is 1, y is 2. By substitution:

$2=-(1)^3+1+c$

Solve for c. Simplify:

$2=-1+1+c$

Then it follows that:

$c=2$

Hence, our entire equation is:

$y=-x^3+x+2$

Part II)

Our derivative is:

$y^\prime=-3x^2+1$

If the gradient is -2, y’ is -2. Hence:

$-2=-3x^2+1$

Solve for x. Subtracting 1 from both sides yields:

$-3=-3x^2$

So:

$x^2=1$

Then it follows that:

$x=\pm1$

Since we already have the point (1, 2) where x is 1, our other point is where x is -1.

Using our equation:

$y=-x^3+x+2$

We can see that our second point is:

$y=-(-1)^3+(-1)+2$

Simplify:

$y=1-1+2=2$

So, our second point where the gradient is -2 is at (-1, 2).