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.
$6 x^{2} + 12 x - 18$
Answer
Solution:
Step:1
To determine the minimum value of a quadratic function $f(x) = ax^2 + bx + c$, we use the formula $x = -\frac{b}{2a}$. For a positive $a$, the function's minimum is at $f\left(-\frac{b}{2a}\right)$.
Step:2
Calculate the $x$-coordinate of the vertex using $x = -\frac{b}{2a}$.
Step:2.1
Insert the given values for $a$ and $b$: $x = -\frac{12}{2(6)}$.
Step:2.2
Simplify the expression by removing the parentheses: $x = -\frac{12}{2(6)}$.
Step:2.3
Further simplify the fraction $-\frac{12}{2(6)}$.
Step:2.3.1
Identify and cancel out common factors between the numerator and denominator.
Step:2.3.1.1
Extract the factor of $2$ from $12$: $x = -\frac{2 \cdot 6}{2 \cdot 6}$.
Step:2.3.1.2
Eliminate the common factors.
Step:2.3.1.2.1
Factor out $2$ from $2 \cdot 6$: $x = -\frac{2 \cdot 6}{2(6)}$.
Step:2.3.1.2.2
Cancel out the common factor: $x = -\frac{\cancel{2} \cdot 6}{\cancel{2} \cdot 6}$.
Step:2.3.1.2.3
Rewrite the simplified expression: $x = -\frac{6}{6}$.
Step:2.3.2
Cancel the common factor of $6$.
Step:2.3.2.1
Perform the cancellation: $x = -\frac{\cancel{6}}{\cancel{6}}$.
Step:2.3.2.2
Express the result: $x = -1$.
Step:2.3.3
Complete the multiplication: $x = -1$.
Step:3
Compute $f(-1)$.
Step:3.1
Substitute $x$ with $-1$ in the function: $f(-1) = 6(-1)^2 + 12(-1) - 18$.
Step:3.2
Simplify the expression.
Step:3.2.1
Simplify each term individually.
Step:3.2.1.1
Square $-1$: $f(-1) = 6 \cdot 1 + 12(-1) - 18$.
Step:3.2.1.2
Multiply $6$ by $1$: $f(-1) = 6 + 12(-1) - 18$.
Step:3.2.1.3
Multiply $12$ by $-1$: $f(-1) = 6 - 12 - 18$.
Step:3.2.2
Combine the numbers by subtraction.
Step:3.2.2.1
Subtract $12$ from $6$: $f(-1) = -6 - 18$.
Step:3.2.2.2
Subtract $18$ from $-6$: $f(-1) = -24$.
Step:3.2.3
The minimum value of the function is $-24$.
Step:4
Determine the coordinates of the minimum point: $(-1, -24)$.
Knowledge Notes:
To find the maximum or minimum value of a quadratic function $f(x) = ax^2 + bx + c$, we use the vertex formula. The vertex $(h, k)$ of the parabola can be found using $h = -\frac{b}{2a}$ and then evaluating $k = f(h)$. If $a > 0$, the parabola opens upwards and the vertex represents the minimum point. If $a < 0$, the parabola opens downwards and the vertex represents the maximum point.
The steps involve:
Identifying the coefficients $a$, $b$, and $c$ from the quadratic equation.
Using the vertex formula to find the $x$-coordinate of the vertex.
Substituting the $x$-coordinate back into the function to find the $y$-coordinate, which gives the maximum or minimum value.
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.
