Given the parabola:
[tex]y=9x^2-36x+37[/tex]Part 1
To determine the way the parabola opens, we consider the coefficient of x².
• If the coefficient is positive, it opens downwards.
,• If the coefficient is negative, it opens upwards.
In this case, the coefficient of x²=9 (Positive).
The parabola opens downwards.
Part 2
The minimum value of the parabola occurs at the line of symmetry.
First, we find the equation of the line of symmetry.
[tex]\begin{gathered} x=-\frac{b}{2a};a=9,b=-36,c=37 \\ \therefore x=-\frac{(-36)}{2\times9} \\ x=2 \end{gathered}[/tex]Find the value of y when x=2.
[tex]\begin{gathered} y=9x^2-36x+37 \\ y=9(2)^2-36(2)+37 \\ =36-72+37 \\ Min\text{imum value of y=1} \end{gathered}[/tex]Part 3
Since the graph has a minimum value, the maximum value of y will be ∞.
Part 5
As obtained in part 2 above, the axis of symmetry is:
[tex]x=2[/tex]Part 6
The vertex is the coordinate of the minimum point.
At the minimum point, when x=2, y=1.
Therefore, the vertex is (2,1).
Part 7
The y-intercept is the value of y when x=0.
[tex]\begin{gathered} y=9x^2-36x+37 \\ y=9(0)^2-36(0)+37 \\ y=37 \end{gathered}[/tex]The y-intercept is 37.
Part 8
We rewrite the equation in Vertex form below:
[tex]\begin{gathered} y=9x^2-36x+37 \\ y-37=9x^2-36x \\ y-37+36=9(x^2-4x+4) \\ y-1=9(x-2)^2 \\ y=9(x-2)^2+1 \end{gathered}[/tex]