Minimum or maximum value of the function
Find the minimum or maximum value of a quadratic function by locating the vertex: use x=-b/2a to find the axis of symmetry, then substitute to get the optimal function value.
Key Concepts
When $a$ is positive in $f(x) = a(x h)^2+k$, the vertex is a low point and its $y$ coordinate is the minimum value of the function . When $a$ is negative, the vertex is a high point and its $y$ coordinate is the maximum value of the function .
In $f(x) = 2(x 3)^2 + 5$, since $a=2$ is positive, the parabola opens up and has a minimum value of 5. In $g(x) = 0.5(x + 1)^2 + 10$, since $a= 0.5$ is negative, it opens down and has a maximum value of 10. A stock's value is modeled by $V(t) = 5(t 4)^2 + 150$. The stock has a maximum value of 150 dollars.
Think of the parabola's mood! If 'a' is positive, the parabola opens upward like a smile, so its vertex is the lowest point (a minimum). If 'a' is negative, it opens downward like a frown, making the vertex the highest point (a maximum). The sign of 'a' tells you if you're looking for a valley or a peak.
Common Questions
How do you determine whether a quadratic function has a minimum or maximum?
The sign of the leading coefficient a determines this. If a>0 the parabola opens upward and the vertex is a minimum. If a<0 the parabola opens downward and the vertex is a maximum.
What is the formula for finding the vertex x-coordinate?
For f(x)=ax^2+bx+c, the x-coordinate of the vertex is x=-b/(2a). Substitute this value back into the function to find the y-coordinate, which is the minimum or maximum value of the function.
How is the vertex used in real-world optimization problems?
In word problems such as maximizing revenue or minimizing cost, the quadratic model's vertex gives the optimal input value and the best possible outcome. For profit P(x)=-2x^2+40x-50, the maximum profit occurs at x=10 units, giving P(10)=$150.