Question 995453
Vertex form is y= a(x-h)^2+k
for a
a(x-3)^2=y, since k=0
But (0,8) is a solution, so y=8 when x=0.
8=-(3)^2*a
9a=8
a=(8/9)
y=(8/9)(x^2-6x+9)
{{{graph(300,300,-10,10,-10,10,(64/81)x^2-(48/9)x+(9))}}}
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minimum of 2 means the y-value of the vertex is 2.
axis of symmetry of 3 means the x value of the vertex is 3.
Therefore, the vertex is at (3,2)
y intercept is at 8.
To draw this, the vertex is at (3,2) and one point is at (0,8).  Use the idea of symmetry to realize when x goes to the left 3, y goes up 6.  When x goes to the right 3,  y goes up 6.  Your third point is (6,8).  With those 3 points, you can draw the parabola.
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y-intercept of 4 means when x=0, y=4.
No zeros means this opens upward.
y=x^2+4.  Note, with this one, y=ax^2+4, where a is positive, will work.