The equation given is:

$y = ax^2 + bx + 1$

We are told that the curve has a gradient of 2 at the point $(1, 0)$.

Step-by-step solution:

1. Differentiate the equation to find the gradient (or slope) function: $\frac{dy}{dx} = 2ax + b$

2. Use the point $(1, 0)$ to determine the gradient at $x = 1$. The gradient is given as 2, so: $2a(1) + b = 2$ Simplifying: $2a + b = 2 \quad \text{(Equation 1)}$

3. Substitute $(1, 0)$ into the original equation $y = ax^2 + bx + 1$ to get a second equation: $0 = a(1)^2 + b(1) + 1$ This simplifies to: $a + b + 1 = 0$ Rearranging: $a + b = -1 \quad \text{(Equation 2)}$

4. Solve the system of equations: From Equation 1: $2a + b = 2$ From Equation 2: $a + b = -1$ Subtract Equation 2 from Equation 1: $(2a + b) - (a + b) = 2 - (-1)$ Simplifying: $a = 3$

5. Substitute $a = 3$ back into Equation 2: $3 + b = -1$ Solving for $b$: $b = -4$

Final Answer:

The values of $a$ and $b$ are: $a = 3, \quad b = -4$

Would you like more details or have any questions about this solution?

Here are five related questions to expand on this topic:

1. What is the significance of the gradient of a curve at a particular point?
2. How would the equation change if the gradient at $x = 1$ were different?
3. Can you find the equation of the tangent line to the curve at the point $(1, 0)$?
4. What happens to the shape of the parabola as the values of $a$ and $b$ change?
5. How would you verify that the values of $a = 3$ and $b = -4$ are correct?

Tip: When solving systems of equations, always double-check the substitutions to avoid calculation errors.