Determine the maximum or minimum value of y-3x2 12x-7 by completing the square?

Answer 1

Every polynomial defines a function, often called P. Linear and and quadratic function belong to a family of functions known as polynomial functions, which often are called P(x). When P(x) = 0, we call it an equation. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. Polynomials of the first few degrees have a special names such as:

Degree 0: Constant function

Degree 1: Linear function

Degree 2: Quadratic function

Degree 3: Cubic function

Degree 4: Quartic function

Degree 5: Quintic function

So, if we work a little bit to the given expression, we can turn it in a polynomial function of the second degree.

y - 3x^2 = 12x - 7

y - 3x^2 + 3x^2 = 12x - 7 + 3x^2

y = 3x^2 + 12x - 7

Let's write y = f(x) and f(x) = 3x^2 + 12x - 7, where a = 3, b = 12, and c = -7.

Since a > 0, the parabola opens upward, so we have a minimum value of the function. The maximum or minimum value of the quadratic function occurs at x = -(b/2a).

x = -12/6 = -2

To find the minimum value of the function, which is also the y-value, we will find f(-2).

f(-2) = 3(-2)^2 + 12(-2) - 7

f(-2) = 12 - 24 - 7 = -19

Thus the minimum value of the function is -19, and the vertex is (-2, -19)

To find zeros, we solve f(x) = 0. So,

f(x) = 3x^2 + 12x - 7

f(x) = 0

3x^2 + 12x - 7 = 0 In order to solve this equation by completing the square, we need the constant term on the right hand side;

3x^2 + 12x = 7 Add the square of one half of the coefficient of x to both sides, (6^2)

3x^2 +12x + 36 = 7 + 36 Use the formula (a + b)^2 = a^2 + 2ab + b^2;

(3x + 6)^2 = 43 Take the square root of both sides, and solve for x;

3x + 6 = (+ & -)square root of 43

3x + 6 = (+ & -)(square root of 43) Subtract 6 to both sides;

3x = (+ & -)(square root of 43) - 6 Divide both sides by 3;

x = (square root of 43)/3 - 2 or x = -(square root of 43)/3 - 2

The solution are (square root of 43) - 2 and -(square root of 43) - 2