In the xycoordinate 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, c = 2 B.) a = 2, b = 6, c = 2 C.) a = 3, b = 3, c = 0 D.) a = 3, b = 6, c = 2
First, find the zeros of the equation. When finding zeros of a quadratic equation, you are finding the xintercepts. At the xintercept, the value of y is 0. To find the xintercepts, substitute 0 for y.
y = 3x²  12x  36 3x²  12x  36 = 0 {substituted 0 for y} 3(x²  4x  12) = 0 {factored out the greatest common factor, 3} 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 {solved each equation for x} It is given that the zeros are at x = a and x = b, and since it is also given that a < b, then a = 2 and b = 6. It is given the graph has a minimum at (c,48). A minimum of a quadratic equation is, graphically, the vertex of the parabola. Since the minimum is at (c,48), then the ycoordinate of the vertex is 48. To find out what x is when y is 48, substitute 48 in for y into the equation and you will be finding the value of c. y = 3x²  12x  36 48 = 3x²  12x  36 {substituted 48 in for y} 3x²  12x + 12 = 0 {added 48 to each side} 3(x²  4x + 4) = 0 {factored out the greatest common factor, 3} 3(x  2)(x  2) = 0 {factored into two binomials} x  2 = 0 {set each factor equal to 0} x = 2 {added 2 to each side} The value of x, when y is 48, is 2. Therefore, the xcoordinate of the vertex is 2, meaning c = 2. B.) a = 2, b = 6, c = 2 Ask Algebra House Comments are closed.

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