In the xy-coordinate plane, the graph of the equation y = 3x² - 12x - 36 has zeros at x = a and x = b,

where a < b. The graph has a minimum at (c , -48). What are the values of a, b, and c?

A.) a = 2, b = 4, and c = 2

B.) a = -2, b = 6, and c = 2

C.) a = -3, b = 3, and c = 0

D.) a = 3, b = 6, and c = 2

where a < b. The graph has a minimum at (c , -48). What are the values of a, b, and c?

A.) a = 2, b = 4, and c = 2

B.) a = -2, b = 6, and c = 2

C.) a = -3, b = 3, and c = 0

D.) a = 3, b = 6, and c = 2

y = 3x² - 12x - 36

3x² - 12x - 36 = 0 {substituted 0 for y}

3(x² - 4x - 12) = 0 {factored out the greatest common factor, 3}

x² - 4x - 12 = 0 {divided each side by 3}

(x - 6)(x + 2) = 0 {factored into two binomials}

x - 6 = 0 or x + 2 = 0 {set each factor equal to 0}

x = 6 or x = -2 {added 6 and subtracted 2, respectively}

Since a < b

a = -2 and b = 6

The minimum, is the y-coordinate of the vertex, -48. To find the the value of the x-coordinate of the

vertex, c, substitute -48 in for y into the equation y = 3x² - 12x - 36 and solve for x.

-48 = 3x² - 12x - 36

3x² - 12x + 12 = 0 {added 48 to each side}

3(x² - 4x + 4) = 0 {factored out the greatest common factor, 3}

x² - 4x + 4 = 0 {divided each side by 3}

(x - 2)(x - 2) = 0 {factored into two binomials}

x - 2 = 0 {set the factor equal to 0}

x = 2 {added 2 to each side}

Therefore, the x-coordinate of the vertex, c, is 2.

This is very poor mathematical questioning on PARCC's behalf. Since we are all taught that standard form of a quadratic equation is ax² + bx + c = 0, and also the x-coordinate of the vertex is -b/2a. Common sense and common mathematical courtesy would tell anyone making an Algebra assessment to use different variables in place of a, b, and c, for this particular question, to avoid confusion. Unless the object is to trick students, which is improper, useless, and intentionally setting them up for failure! This is not surprising.

3x² - 12x - 36 = 0 {substituted 0 for y}

3(x² - 4x - 12) = 0 {factored out the greatest common factor, 3}

x² - 4x - 12 = 0 {divided each side by 3}

(x - 6)(x + 2) = 0 {factored into two binomials}

x - 6 = 0 or x + 2 = 0 {set each factor equal to 0}

x = 6 or x = -2 {added 6 and subtracted 2, respectively}

Since a < b

a = -2 and b = 6

The minimum, is the y-coordinate of the vertex, -48. To find the the value of the x-coordinate of the

vertex, c, substitute -48 in for y into the equation y = 3x² - 12x - 36 and solve for x.

-48 = 3x² - 12x - 36

3x² - 12x + 12 = 0 {added 48 to each side}

3(x² - 4x + 4) = 0 {factored out the greatest common factor, 3}

x² - 4x + 4 = 0 {divided each side by 3}

(x - 2)(x - 2) = 0 {factored into two binomials}

x - 2 = 0 {set the factor equal to 0}

x = 2 {added 2 to each side}

Therefore, the x-coordinate of the vertex, c, is 2.

**B.) a = -2, b = 6, and c = 2**__Commentary from www.algebrahouse.com:__This is very poor mathematical questioning on PARCC's behalf. Since we are all taught that standard form of a quadratic equation is ax² + bx + c = 0, and also the x-coordinate of the vertex is -b/2a. Common sense and common mathematical courtesy would tell anyone making an Algebra assessment to use different variables in place of a, b, and c, for this particular question, to avoid confusion. Unless the object is to trick students, which is improper, useless, and intentionally setting them up for failure! This is not surprising.

**- Algebra House**