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.
f(x)=4-x^2
⇕
f(x)= - 1x^2+ 0 * x+ 4
We can see above that a= - 1, b= 0, and c= 4. We will now use these values to find the desired information.
Maximum Value
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= 0 in - b2a.
This tells us that the maximum value of the function is 4.
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, 4.
Domain:& All real numbers
Range:& y ≤ 4
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 0.
Increasing Interval:& To the left of 0
Decreasing Interval:& To the right of 0
