AP Calculus AB Exam Question:

Consider the curve defined by the equation y = 3x^2 - 5x + 2 for -1 ≤ x ≤ 2.

a) Find the exact coordinates of the points where the curve intersects the x-axis.

b) Find the coordinates of the vertex of the parabola.

c) Find the area of the region enclosed by the curve and the x-axis.

Answer:

a) To find the points where the curve intersects the x-axis, we need to find the values of x for which y equals zero. Thus, we solve the equation 3x^2 - 5x + 2 = 0.

Factoring the quadratic equation, we have:

(3x - 1)(x - 2) = 0

Setting each factor equal to zero, we find:

3x - 1 = 0 --> x = 1/3 x - 2 = 0 --> x = 2

Therefore, the curve intersects the x-axis at the points (1/3, 0) and (2, 0).

b) To find the coordinates of the vertex of the parabola, we use the formula x = -b/(2a), where a and b are the coefficients of the quadratic equation in the form ax^2 + bx + c.

In our case, a = 3 and b = -5. Substituting these values into the formula, we have:

x = -(-5) / (2*3) = 5/6

To find the corresponding value of y, we substitute this x-value into the original equation:

y = 3(5/6)^2 - 5(5/6) + 2

Simplifying the expression, we get:

y = 5/4 - 25/6 + 2 = -7/12

Therefore, the vertex of the parabola is at the coordinates (5/6, -7/12).

c) To find the area of the region enclosed by the curve and the x-axis, we need to calculate the definite integral of the function y = 3x^2 - 5x + 2 over the interval -1 to 2.

∫[from -1 to 2] (3x^2 - 5x + 2) dx

First, let's find the antiderivative of each term:

∫(3x^2) dx = x^3 + C1 ∫(-5x) dx = -5/2 * x^2 + C2 ∫(2) dx = 2x + C3

Now, integrating each term, we have:

= [x^3 + C1] [from -1 to 2] + [-5/2 * x^2 + C2] [from -1 to 2] + [2x + C3] [from -1 to 2]

= [(2^3 + C1) - (-(-1)^3 + C1)] + [-5/2 * (2^2) + C2 - (-5/2 * (-1)^2 + C2)] + [2(2) + C3 - (2(-1) + C3)]

= [8 + C1 - 1 + C1] + [-20/2 + C2 - (-5/2 + C2)] + [4 + C3 - (-2 + C3)]

= 16 + 2C1 + 5 + 2C2 + 6 + 2C3

= 27 + 2C1 + 2C2 + 2C3

Since this is a definite integral, the constant terms cancel out, and we are left with:

Area = 27

Therefore, the area of the region enclosed by the curve and the x-axis is 27 square units.