Answer :
To solve for the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we will follow a few steps involving the given conditions. Let's break it down:
### Step 1: Understanding the Gradient at a Point
The gradient of the curve at any point [tex]\(x\)[/tex] is given as:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
### Step 2: Condition at [tex]\( (2, -16) \)[/tex]
The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis. This implies that the gradient at [tex]\(x = 2\)[/tex] is 0. Therefore:
[tex]\[ \left. \frac{dy}{dx} \right|_{x = 2} = 0 \][/tex]
Substituting [tex]\(x = 2\)[/tex] into the gradient expression:
[tex]\[ a (2)^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
So, we have our first equation:
[tex]\[ b = -4a \][/tex]
### Step 3: Integrating the Gradient
To find the curve [tex]\(y\)[/tex] which passes through the given points, we need to integrate the gradient function:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
Integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = \frac{a}{3} x^3 + bx + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 4: Applying the Given Point [tex]\((1, -11)\)[/tex]
The curve passes through the point [tex]\((1, -11)\)[/tex], so we substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation of the curve:
[tex]\[ -11 = \frac{a}{3} (1)^3 + b(1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
Using [tex]\( b = -4a \)[/tex], substitute [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a}{3} - \frac{12a}{3} + C \][/tex]
[tex]\[ -11 = -\frac{11a}{3} + C \][/tex]
Multiplying through by 3:
[tex]\[ -33 = -11a + 3C \][/tex]
[tex]\[ 3C = -33 + 11a \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
### Step 5: Applying the Given Point [tex]\((2, -16)\)[/tex]
The curve also passes through the point [tex]\((2, -16)\)[/tex], so we substitute [tex]\(x = 2\)[/tex] and [tex]\(y = -16\)[/tex] into the equation of the curve:
[tex]\[ -16 = \frac{a}{3} (2)^3 + b(2) + C \][/tex]
[tex]\[ -16 = \frac{8a}{3} + 2b + C \][/tex]
Using [tex]\( b = -4a \)[/tex] and [tex]\( C = -11 + \frac{11a}{3}\)[/tex], substitute [tex]\(b\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ -16 = \frac{8a}{3} + 2(-4a) + \left( -11 + \frac{11a}{3} \right) \][/tex]
[tex]\[ -16 = \frac{8a}{3} - 8a - 11 + \frac{11a}{3} \][/tex]
[tex]\[ -16 = \frac{8a + 11a}{3} - 8a - 11 \][/tex]
[tex]\[ -16 = \frac{19a}{3} - 8a - 11 \][/tex]
Combine like terms:
[tex]\[ -16 = \frac{19a}{3} - \frac{24a}{3} - 11 \][/tex]
[tex]\[ -16 = -\frac{5a}{3} - 11 \][/tex]
[tex]\[ -16 + 11 = -\frac{5a}{3} \][/tex]
[tex]\[ -5 = -\frac{5a}{3} \][/tex]
[tex]\[ 5 = \frac{5a}{3} \][/tex]
Multiply both sides by 3:
[tex]\[ 15 = 5a \][/tex]
[tex]\[ a = 3 \][/tex]
Now substitute [tex]\(a = 3\)[/tex] back into the equation for [tex]\(b\)[/tex]:
[tex]\[ b = -4a \][/tex]
[tex]\[ b = -4(3) = -12 \][/tex]
### Conclusion
The values of the constants are:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = -12 \][/tex]
### Step 1: Understanding the Gradient at a Point
The gradient of the curve at any point [tex]\(x\)[/tex] is given as:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
### Step 2: Condition at [tex]\( (2, -16) \)[/tex]
The tangent to the curve at the point [tex]\((2, -16)\)[/tex] is parallel to the [tex]\(x\)[/tex]-axis. This implies that the gradient at [tex]\(x = 2\)[/tex] is 0. Therefore:
[tex]\[ \left. \frac{dy}{dx} \right|_{x = 2} = 0 \][/tex]
Substituting [tex]\(x = 2\)[/tex] into the gradient expression:
[tex]\[ a (2)^2 + b = 0 \][/tex]
[tex]\[ 4a + b = 0 \][/tex]
So, we have our first equation:
[tex]\[ b = -4a \][/tex]
### Step 3: Integrating the Gradient
To find the curve [tex]\(y\)[/tex] which passes through the given points, we need to integrate the gradient function:
[tex]\[ \frac{dy}{dx} = a x^2 + b \][/tex]
Integrate with respect to [tex]\(x\)[/tex]:
[tex]\[ y = \int (a x^2 + b) \, dx \][/tex]
[tex]\[ y = \int a x^2 \, dx + \int b \, dx \][/tex]
[tex]\[ y = \frac{a}{3} x^3 + bx + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 4: Applying the Given Point [tex]\((1, -11)\)[/tex]
The curve passes through the point [tex]\((1, -11)\)[/tex], so we substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -11\)[/tex] into the equation of the curve:
[tex]\[ -11 = \frac{a}{3} (1)^3 + b(1) + C \][/tex]
[tex]\[ -11 = \frac{a}{3} + b + C \][/tex]
Using [tex]\( b = -4a \)[/tex], substitute [tex]\(b\)[/tex] into the equation:
[tex]\[ -11 = \frac{a}{3} - 4a + C \][/tex]
[tex]\[ -11 = \frac{a}{3} - \frac{12a}{3} + C \][/tex]
[tex]\[ -11 = -\frac{11a}{3} + C \][/tex]
Multiplying through by 3:
[tex]\[ -33 = -11a + 3C \][/tex]
[tex]\[ 3C = -33 + 11a \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
[tex]\[ C = -11 + \frac{11a}{3} \][/tex]
### Step 5: Applying the Given Point [tex]\((2, -16)\)[/tex]
The curve also passes through the point [tex]\((2, -16)\)[/tex], so we substitute [tex]\(x = 2\)[/tex] and [tex]\(y = -16\)[/tex] into the equation of the curve:
[tex]\[ -16 = \frac{a}{3} (2)^3 + b(2) + C \][/tex]
[tex]\[ -16 = \frac{8a}{3} + 2b + C \][/tex]
Using [tex]\( b = -4a \)[/tex] and [tex]\( C = -11 + \frac{11a}{3}\)[/tex], substitute [tex]\(b\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ -16 = \frac{8a}{3} + 2(-4a) + \left( -11 + \frac{11a}{3} \right) \][/tex]
[tex]\[ -16 = \frac{8a}{3} - 8a - 11 + \frac{11a}{3} \][/tex]
[tex]\[ -16 = \frac{8a + 11a}{3} - 8a - 11 \][/tex]
[tex]\[ -16 = \frac{19a}{3} - 8a - 11 \][/tex]
Combine like terms:
[tex]\[ -16 = \frac{19a}{3} - \frac{24a}{3} - 11 \][/tex]
[tex]\[ -16 = -\frac{5a}{3} - 11 \][/tex]
[tex]\[ -16 + 11 = -\frac{5a}{3} \][/tex]
[tex]\[ -5 = -\frac{5a}{3} \][/tex]
[tex]\[ 5 = \frac{5a}{3} \][/tex]
Multiply both sides by 3:
[tex]\[ 15 = 5a \][/tex]
[tex]\[ a = 3 \][/tex]
Now substitute [tex]\(a = 3\)[/tex] back into the equation for [tex]\(b\)[/tex]:
[tex]\[ b = -4a \][/tex]
[tex]\[ b = -4(3) = -12 \][/tex]
### Conclusion
The values of the constants are:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = -12 \][/tex]
