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

and c are constants. The gradient of the curve at the
point (1, 2) is -2.
(i) Find the value of a and of c.
(ii) Hence, find the coordinates of the other point
on the curve where the gradient is -2.​

User Xhantar
by
6.3k points

1 Answer

6 votes

Answer:

Part I)


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

Giving the equation:


y=-x^3+x+2

Part II)

(-1, 2)

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=(d)/(dx)\big[ax^3+x+c\big]

Expand the right:


\displaystyle y^\prime=(d)/(dx)\big[ax^3]+(d)/(dx)\big[x\big]+(d)/(dx)\big[c\big]

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


\displaystyle y^\prime=a(d)/(dx)\big[x^3]+(d)/(dx)\big[x\big]+c(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).

User TheRealWorld
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6.9k points