Answer:
Part 1) The vertex is the point (1,-9)
Part 2) The equation of the axis of symmetry is x=1
Part 3) The y-intercept is the point (0,-8)
Part 4) The x-intercepts are (-2,0) and (4,0)
Part 5) The graph in the attached figure
Step-by-step explanation:
we have
[tex]h(x)=(x-1)^2-9[/tex]
This is a vertical parabola open upward
The vertex represent a minimum
Part 1) Find the vertex
The quadratic equation is written in vertex form
[tex]y=a(x-h)^2+k[/tex]
where
(h,k) is the vertex of the parabola
so
The vertex is the point (1,-9)
Part 2) Find the axis of symmetry
The equation of the axis of symmetry of a vertical parabola is equal the the x-coordinate of the vertex
so
The equation of the axis of symmetry is
[tex]x=1[/tex]
Part 3) Find the y-intercept
we know that
The y-intercept is the value of the function when the value of x is equal to zero
so
For x=0
[tex]h(x)=(0-1)^2-9[/tex]
[tex]h(0)=-8[/tex]
therefore
The y-intercept is the point (0,-8)
Part 4) Find the x-intercepts
we know that
The x-intercepts are the values of x when the value of the function is equal to zero
so
For h(x)=0
[tex](x-1)^2-9=0[/tex]
solve for x
[tex](x-1)^2=9[/tex]
square root both sides
[tex]x-1=\pm3[/tex]
[tex]x=1\pm3[/tex]
[tex]x=1+3=4\\x=1-3=-2[/tex]
therefore
The x-intercepts are (-2,0) and (4,0)
Part 5) Plot the graph of the quadratic function
Plot the following points to graph the function
vertex (1,-9)
axis of symmetry x=1
y-intercept (0,-8)
x-intercepts (-2,0) and (4,0)
The graph in the attached figure
