Question 957325: A quadratic relation has the equation y=a(x-s)(x-t)
Find the value of a when
a) y=a(x-2)(x+6) and (3,5) is a point on the graph
b)the parabola has zeros of 4 and -2 and a y-intercept of 1
c)the parabola has x-intercepts of 4 and -2 and a y-intercept of -1
d)the parabola has zeros of 5 and 0 and a minimum value of -10
e)the parabola has x-intercepts of 5 and -3 and a maximum value of 6
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A quadratic relation has the equation y=a(x-s)(x-t)
Find the value of a when
a) y=a(x-2)(x+6) and (3,5) is a point on the graph
5 = a(3-2)(3+6)
5 = a*1*9
a = 5/9
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b)the parabola has zeros of 4 and -2 and a y-intercept of 1
y = a(x-4)(x+2)
1 = a(-4)(2)
a = -1/8
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c)the parabola has x-intercepts of 4 and -2 and a y-intercept of -1
y = a(x-4)(x+2)
-1 = a(-4)(2)
a = 1/8
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d)the parabola has zeros of 5 and 0 and a minimum value of -10
y = a(x-5)(x)
Note: The vertex is midway between 5 and 0: i.e. x = 5/2
10 = a(5/2 -5)(5/2)
10 = a(-5/2)(5/2)
10 = a(-25/4)
a = -40/25
a = -8/5
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e)the parabola has x-intercepts of 5 and -3 and a maximum value of 6
y = a(x-5)(x+3)
Max at (1,6)
6 = a(-4)(4)
a = -6/16 = -3/8
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cheers,
Stan H.
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