Find the Maximum/Minimum Value 6x^2+12x-18
The given problem requests to determine the highest (maximum) or lowest (minimum) possible value of the quadratic function 6x^2 + 12x - 18 over its domain (which is all real numbers). This function is a parabola, and depending on the direction it opens (which is determined by the coefficient of the x^2 term), the parabola will have either a minimum point (if it opens upwards) or a maximum point (if it opens downwards) at its vertex. The question involves finding this extremum value by possibly completing the square, using calculus, or applying the vertex-formula for quadratic functions.
To determine the minimum value of a quadratic function
Calculate the
Insert the given values for
Simplify the expression by removing the parentheses:
Further simplify the fraction
Identify and cancel out common factors between the numerator and denominator.
Extract the factor of
Eliminate the common factors.
Factor out
Cancel out the common factor:
Rewrite the simplified expression:
Cancel the common factor of
Perform the cancellation:
Express the result:
Complete the multiplication:
Compute
Substitute
Simplify the expression.
Simplify each term individually.
Square
Multiply
Multiply
Combine the numbers by subtraction.
Subtract
Subtract
The minimum value of the function is
Determine the coordinates of the minimum point:
To find the maximum or minimum value of a quadratic function
The steps involve:
Identifying the coefficients
Using the vertex formula to find the
Substituting the
Simplifying expressions and performing arithmetic operations as needed.
Interpreting the result in the context of the problem, such as finding the maximum height, minimum cost, etc.