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for the quadratic function ( f(x)=-x^{2}-8x ), answer parts (a) through…

Question

for the quadratic function ( f(x)=-x^{2}-8x ), answer parts (a) through (f)
a. the ( y )-intercept is 0
(type an integer or a simplified fraction.)
b. there is no ( y )-intercept.
what is the ( x )-intercept? select the correct choice below and, if necessary, fill in the answer box
complete your choice
a. the ( x )-intercept(s) is/are ( -8,0 )
(type an integer or a simplified fraction. use a comma to separate answers as needed )
b. there is/are no ( x )-intercept(s)
(c) use parts (a) and (b) to graph the function.
use the graphing tool to graph the function.
(d) find the domain and the range of the quadratic function.
the domain of ( f ) is ( (-infty,infty) ).
(type your answer in interval notation.)
the range of ( f ) is ( (-infty,16 ).
(type your answer in interval notation )
(e) determine where the quadratic function is increasing and where it is decreasing.
the function is increasing on the interval ( (-infty,-4) ).
(type your answer in interval notation.)
the function is decreasing on the interval
(type your answer in interval notation.)

Explanation:

Step1: Find the vertex of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\)-coordinate of the vertex is \(x=-\frac{b}{2a}\).
For \(f(x)=-x^{2}-8x\), where \(a=-1\) and \(b = - 8\), then \(x=-\frac{-8}{2\times(-1)}=-4\).
Substitute \(x = - 4\) into \(f(x)\): \(f(-4)=-(-4)^{2}-8\times(-4)=-16 + 32=16\).
The vertex is \((-4,16)\). Since \(a=-1<0\), the parabola opens down - ward.

Step2: Determine the decreasing interval

A parabola \(y = ax^{2}+bx + c\) (\(a<0\)) is increasing on \((-\infty,-\frac{b}{2a})\) and decreasing on \((-\frac{b}{2a},\infty)\)
Since \(x =-\frac{b}{2a}=-4\) and \(a=-1<0\)

Answer:

The function is decreasing on the interval \((-4,\infty)\)