Find the range of the function \(y= 2x^2 - 3x + 1\) when \(x\) is in the interval \([-2,3]\).
Answer
Step 4: The maximum value of the function in the interval \([-2,3]\) is the larger one among the function values at the endpoints of the interval. Substituting \(x = -2\) and \(x = 3\) into the function, we get \(2(-2)^2 - 3(-2) + 1 = 11\) and \(2(3)^2 - 3(3) + 1 = 10\), respectively, so the maximum value is 11.
Step 1 ：Step 1: Since the function \(y= 2x^2 - 3x + 1\) is a quadratic function and the coefficient of \(x^2\) is positive, the graph of the function opens upwards. This means the function has a minimum value.
Step 2 ：Step 2: The minimum value of the function occurs at the vertex of the parabola. The \(x\)-coordinate of the vertex of a parabola \(y = ax^2 + bx + c\) is given by \(-\frac{b}{2a}\). Substituting \(a = 2\) and \(b = -3\) into this formula, we get \(-\frac{-3}{2\times2} = 0.75\).
Step 3 ：Step 3: Since \(0.75\) is in the interval \([-2,3]\), the minimum value of the function in this interval is the function value at \(x = 0.75\), which is \(2(0.75)^2 - 3(0.75) + 1 = 0.125\).
Step 4 ：Step 4: The maximum value of the function in the interval \([-2,3]\) is the larger one among the function values at the endpoints of the interval. Substituting \(x = -2\) and \(x = 3\) into the function, we get \(2(-2)^2 - 3(-2) + 1 = 11\) and \(2(3)^2 - 3(3) + 1 = 10\), respectively, so the maximum value is 11.
