Quadratic function question

wilc0919

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Oct 13, 2006
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37
Suppose that a quadratic function has it's vertex at (3,0) and has y-intercept of -4. Find a formula for the function....

I got

y= -4/9x^2 + 24/9x -4

is that correct???
 
When x = 3, you get:

. . . . .-(4/9)[3]<sup>2</sup> + (24/9)[3] - 4

. . . . .-(4/9)[9] + (24/3) - 4

. . . . .-4 + 8 - 4

. . . . .0

...so the point (x, y) = (3, 0) is on this curve. Checking for the vertex:

. . . . .h = -b/(2a)

. . . . .= -(24/9) / (2(-4/9))

. . . . .= (-8/3) / (-8/9)

. . . . .= (-8/3)(-9/8)

. . . . .= 3

...so the point (3, 0) is the vertex, (h, k). Your equation would appear to be correct.

Eliz.
 
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