Start by identifying a, b, and c. The maximum value of the given quadratic function is f ( - b2a ).
Maximum Value: 2 Domain: All real numbers Range: y ≤ 2 Increasing Interval: To the left of x=- 2 Decreasing Interval: To the right of x=- 2
Practice makes perfect
For the quadratic function f(x)=ax^2+bx+c, the y-coordinate of the vertex is the maximum value of the function when a<0.
Let's identify the values of a, b, and c in the given quadratic function.
y=- x^2-4x-2
⇕
y= - 1x^2+( - 4)x+( - 2)
We can see above that a= - 1, b= - 4, and c= - 2. We will now use these values to find the desired information.
Maximum Value
We need to think of y as a function of x, y=f(x). Since a= - 1 is less than 0, the parabola will open downwards. This means it will have a maximum value, which is given by f ( - b2a ). Before we find the value of the function at this point, we need to substitute a= - 1 and b= - 4 in - b2a.
This tells us that the maximum value of the function is 2.
Domain and Range
Unless there are any specified restrictions on the x-values, the domain of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since a= - 1 is less than 0, the range is all values less than or equal to the maximum value, 2.
Domain:& All real numbers
Range:& y ≤ 2
Decreasing and Increasing Intervals
Since a= - 1 is less than 0, the function increases to the left of the maximum value and decreases to the right of the maximum value, which we know occurs at - 2.
Increasing Interval:& To the left of - 2
Decreasing Interval:& To the right of - 2
