The function is
[tex]h(x)=(x-1)^2-9[/tex]1) x-intercept(s)
The x-intercepts refer to the points on which the function intercepts with the x-axis, in other words, when y=h(x)=0
So, given that condition, we get
[tex]\begin{gathered} h(x)=0 \\ \Rightarrow(x-1)^2-9=0 \\ \Rightarrow x^2-2x+1^{}-9=0 \\ \Rightarrow x^2-2x-8=0 \\ \Rightarrow(x-4)(x+2)=0 \end{gathered}[/tex]Therefore, there are two x-intercepts, and those are the points
[tex](4,0),(-2,0)[/tex]2) y-intercepts
The y-intercepts happen when x=0. So,
[tex]\begin{gathered} x=0 \\ \Rightarrow h(0)=(0-1)^2-9=1-9=-8 \end{gathered}[/tex]So, there is only one y-intercept and it's on the point (0,-8)
3) Vertex
The general equation of a parabola is
[tex]y=f(x)=a^{}x^2+bx+c[/tex]There is another way to express the same function, which is called the 'vertex form':
[tex]\begin{gathered} y=f(x)=a(x-h)^2+k \\ \Rightarrow y=ax^2-2ahx+ah^2+k \end{gathered}[/tex]What is particularly useful of this vertex form is that the vertex is the point (h,k)
So, transforming h(x) into vertex form:
[tex]\begin{gathered} h(x)=(x-1)^2-9=a(x-h)^2+k \\ \Rightarrow\begin{cases}a=1 \\ h=1 \\ k=-9\end{cases} \end{gathered}[/tex]Therefore, the vertex is the point (h,k)=(1,-9)
4) Axis of symmetry
In general, the equation of the axis of symmetry is given by
[tex]x=-\frac{b}{2a};y=f(x)=ax^2+bx+c[/tex]Therefore, in our particular problem,
[tex]\begin{gathered} h(x)=x^2-2x-8=ax^2+bx+c \\ \Rightarrow\begin{cases}a=1 \\ b=-2 \\ c=-8\end{cases} \\ \end{gathered}[/tex]Thus, the equation of the line that is the axis of symmetry is
[tex]x=-\frac{b}{2a}=-\frac{(-2)}{2\cdot1}=-\frac{(-2)}{2}=1[/tex]Then, the axis of symmetry is the line x=1.
Summing up the information in the four previous steps, we get
