Answer:
Part 1) The domain of the quadratic function is the interval (-∞,∞)
Part 2) The range is the interval (-∞,1]
Step-by-step explanation:
we have
![f(x)=-x^2+14x-48](https://tex.z-dn.net/?f=f%28x%29%3D-x%5E2%2B14x-48)
This is a quadratic equation (vertical parabola) open downward (the leading coefficient is negative)
step 1
Find the domain
The domain of a function is the set of all possible values of x
The domain of the quadratic function is the interval
(-∞,∞)
All real numbers
step 2
Find the range
The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.
we have a vertical parabola open downward
The vertex is a maximum
Let
(h,k) the vertex of the parabola
so
The range is the interval
(-∞,k]
Find the vertex
![f(x)=-x^2+14x-48](https://tex.z-dn.net/?f=f%28x%29%3D-x%5E2%2B14x-48)
Factor -1 the leading coefficient
![f(x)=-(x^2-14x)-48](https://tex.z-dn.net/?f=f%28x%29%3D-%28x%5E2-14x%29-48)
Complete the square
![f(x)=-(x^2-14x+49)-48+49](https://tex.z-dn.net/?f=f%28x%29%3D-%28x%5E2-14x%2B49%29-48%2B49)
![f(x)=-(x^2-14x+49)+1](https://tex.z-dn.net/?f=f%28x%29%3D-%28x%5E2-14x%2B49%29%2B1)
Rewrite as perfect squares
![f(x)=-(x-7)^2+1](https://tex.z-dn.net/?f=f%28x%29%3D-%28x-7%29%5E2%2B1)
The vertex is the point (7,1)
therefore
The range is the interval
(-∞,1]