f is a polynomial function of degree 2. it is a parabola.
for any real value, we can evaluate -x^2-10x therefore
its domain is R
the y-intercept is by definition f(0)
x-intercept are the solution of f(x)=0
iff x=0 or x=-10
x-intercept are 0 and -10
The vertex of a parabola of the for ax^2+bx+c is given at
in our case x=10/2=-5
vertex x=-5 y=25
the coeeficient in front of x^2 is negative therefore the parabola open down.
f is increasing before the vertex
f is decreasing after the vertex
The symmetry axis is x=-5
The parabola is open down therefore its range if
You need to remember that vertex is an extreme point to the function, hence, you should find the critical value of f(x) by differentiation such that:
`f'(x) = -2x - 10`
You need to solve the equation `f'(x)=0 ` so find the critical value such that:
You need to divide by `-2` both sides such that:
`x + 5 = 0 =gt x = -5`
You should substitute `-5` for`x` in equation of function such that:
`f(-5)=-25 + 50 =gt f(-5)=25`
Hence, the vertex of parabola `f(x)=-x^2-10x is at (-5,25).`
You need to remember that the first derivative tells you what are the intervals where the function increases or decreases, hence, substituting a larger value than -5 for x yields a positive amount and substituting a smaller value than -5 for x yields a negative amount, thus the function increases for values larger than -5 and it decreases for values smaller than -5.
Notice that the function is polynomial, hence the domain is the real set and the range consists of negative values only.
The axis of symmetry passes through the vertex of parabola and it is parallel to y axis, hence `x=-5` .
Sketching the graph of parabola, you should notice all the above such that:
Hence, the domain of function is R, the range is interval `(-oo,0],` the axis of symmetry is x=-5, the vertex of parabola is `(-5,25) ` and the function increases over`(-5,+oo)` and it decreases over `(-oo,-5).`