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A curve passes through the point [tex]\((1, -11)\)[/tex] and its gradient at any point is [tex]\(a x^2 + b\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants. The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis.

Find:

i) The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]

ii) The equation of the curve

Sagot :

GinnyAnswer
Let's solve this problem step-by-step.

### i) Finding the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]

Given:
1. The curve passes through the point [tex]\((1, -11)\)[/tex].
2. The gradient at any point on the curve is [tex]\(a x^2 + b\)[/tex].
3. The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis.

Since the tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis, the gradient at [tex]\((2, -16)\)[/tex] must be [tex]\(0\)[/tex]. Therefore, we have:
[tex]\[ a \cdot 2^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
[tex]\[ b = -4a \quad \text{(Equation 1)} \][/tex]

### ii) Finding the equation of the curve

The gradient of the curve (the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex]) is given by:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]

To find the equation of the curve, we need to integrate the gradient function. Integrating [tex]\(a x^2 + b\)[/tex] with respect to [tex]\(x\)[/tex], we get:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = a \int x^2 \, dx + b \int 1 \, dx \][/tex]
[tex]\[ y = a \left( \frac{x^3}{3} \right) + b x + C \][/tex]
[tex]\[ y = \frac{a x^3}{3} + b x + C \][/tex]

Now, we'll use the point [tex]\((1, -11)\)[/tex] to find the constant [tex]\(C\)[/tex]. Substituting [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation:
[tex]\[ -11 = \frac{a (1)^3}{3} + b (1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]

We already have [tex]\(b = -4a\)[/tex] from Equation 1. Substituting [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a - 12a}{3} + C \][/tex]
[tex]\[ -11 = \frac{-11a}{3} + C \][/tex]
[tex]\[ -11a = -33 - 3C \][/tex]
[tex]\[ 3 = 33 + 3C \][/tex]

Now we solve for [tex]\(C\)[/tex]:
[tex]\[ 3C = -22 \][/tex]
[tex]\[ C = -22/3 \][/tex]

### Summarizing the findings

1. From [tex]\(b = -4a\)[/tex]:
- a = b / (-4)
- The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are defined parametrically:
- [tex]\(a = b/( -4 \)[/tex]
- [tex]\(b = -4a\)[/tex]
[tex]\(\therefore\)[/tex] [tex]\[ a = 1/4 \][/tex]

2. The equation of the curve is
\[ y = \frac{1}{4}x^3 - 4(-11) * x + \frac{4}{{-11}} / 13 +23/5.
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