Start by identifying a, b, and c. The minimum value of the given quadratic function is f ( - b2a ).
Minimum Value: - 2 Domain: All real numbers Range: y ≥ - 2 Decreasing Interval: To the left of x=- 2 Increasing Interval: To the right of x=- 2
Practice makes perfect
For a quadratic function f(x)=ax^2+bx+c, the y-coordinate of the vertex is the minimum value of the function when a>0.
Let's identify the values of a, b, and c in the given quadratic function.
f(x)=3/2x^2+6x+4
⇕
f(x)= 3/2x^2+ 6x+ 4
We can see above that a= 32, b= 6, and c= 4. We will now use these values to find the desired information.
Minimum Value
Since a= 32 is greater than 0, the parabola will open upwards. This means it will have a minimum value, which is given by f ( - b2a ). Before we find the value of the function at this point, we need to substitute a= 32 and b= 6 in - b2a.
This tells us that the minimum 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= 32 is greater than 0, the range is all values greater than or equal to the minimum value, - 2.
Domain:& All real numbers
Range:& y ≥ - 2
Decreasing and Increasing Intervals
Since a= 32 is greater than 0, the function decreases to the left of the minimum value and increases to the right of the minimum value, which we know occurs at x=- 2.
Decreasing Interval:& To the left of - 2
Increasing Interval:& To the right of - 2
