A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve.

We are given that the curve passes through the points \((-2, 0)\) and \((3, 0)\). Since the curve is a function of \(x\), we can assume that the equation of the curve is of the form \(y = f(x)\). If the curve passes through the point \((-2, 0)\), then we can substitute \(x = -2\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(-2) $$ Similarly, if the curve passes through the point \((3, 0)\), then we can substitute \(x = 3\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(3) $$ Therefore, the curve must have at least two roots, namely \(x = -2\) and \(x = 3\). This suggests that the curve is a quadratic function. We can use the information about the roots of the curve to write the equation of the curve in factored form: $$ y = A(x + 2)(x - 3) $$ where \(A\) is a constant. Since the coefficient of \(x^2\) in the equation is 1, we can simplify this equation to: $$ y = Ax^2 + Bx + C $$ where \(A = 1\), and \(B\) and \(C\) are constants. To find the values of \(B\) and \(C\), we can substitute the points \((-2, 0)\) and \((3, 0)\) into the equation: $$ 0 = A(-2)^2 + B(-2) + C \quad \text{and} \quad 0 = A(3)^2 + B(3) + C $$ Simplifying these equations, we get: $$ 4A - 2B + C = 0 \quad \text{and} \quad 9A + 3B + C = 0 $$ We can use these two equations to solve for \(B\) and \(C\). Adding the two equations, we get: $$ 13A + B = 0 $$ Substituting this expression for \(B\) into one of the previous equations, we get: $$ 4A - 2(-13A) + C = 0 $$ Simplifying, we get: $$ 20A + C = 0 $$ Solving these two equations simultaneously, we get: $$ A = \frac{1}{13}, \quad B = -\frac{4}{13}, \quad C = 0 $$ Therefore, the equation of the curve is: $$ y = \frac{1}{13}(x + 2)(x - 3) = \frac{1}{13}(x^2 - x - 6) $$ So the correct answer is \(y = x^2 - x - 6\).