Answer:
A: Nature of root
we have a formula to find the roots of quadratic equation
[tex]\frac{-b\pm\sqrt{D}}{2a}[/tex]
where, [tex]D=b^2-4ac[/tex]
We have general form of quadratic equation as
[tex]ax^2+bx+c[/tex]
Here, a=-3,b=6 and c=17 on substituting the values in [tex]D=b^2-4ac[/tex] and [tex]\frac{b\pm\sqrt{D}}{2a}[/tex]
[tex]D=6^2-4(-3)(17)=240[/tex]
Now, substitute D we get
[tex]\frac{-6\pm\sqrt{240}}{2(-3)}=\frac{3+2\sqrt{15}}{3},\frac{3-2sqrt{15}}{3}[/tex]
B: Open upward or downward
This given function is a downward parabola you can refer the figure in the attachment
C: End behaviour
Here, we have negative leading coefficient and even degree
Hence,
[tex]x\rightarrow-\infty[/tex] [tex]f(x)\rightarrow-\infty[/tex]
[tex]x\rightarrow\infty[/tex] [tex]f(x)\rightarrow-\infty[/tex]
A= -3
B= 6
C= 17
Axis of Symmetry: X=1
Vertex (1, 20)
The Vertex is at a: Maximum
