Question 995453: 1)Draw the parabola with a zero at (3,0) ,a vertex at (3,0) and a y intercept of (0,8)
2)Draw the parabola with a minimum value of 2, an axis of symmetry of 3, no zeros, and a y intercept of 8.
3)Draw the parabola with a y intercept of 4, an axis of symmetry of 0, and no zeros.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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)
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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.
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