Find the zeros and x-coordinate of the vertex

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, 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 x-intercepts.  At the x-intercept, the value of y is 0.  To find the x-intercepts, 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 y-coordinate 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 x-coordinate of the vertex is 2, meaning c = 2.

B.)  a = -2, b = 6, c = 2

Ask Algebra House