SOLUTION: write the equation y=a(x-h)^2+k with the given. with a y-intercept 10, x-intercept 2, and equation of axis x-3=0

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Question 274307: write the equation y=a(x-h)^2+k with the given.
with a y-intercept 10, x-intercept 2, and equation of axis x-3=0
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
write the equation y=a(x-h)^2+k with the given.
with a y-intercept 10, x-intercept 2, and equation of axis x-3=0
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You have three points: (0,10), (2,0) and vertex(3,y)
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If vertex is (3,y), h=3.
y = a(x-3)^2+k
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Substituting (0,10) and (2,0)
10 = a(-3)^2 + k
0 = a(-1)^2 + k
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Solve for a and k:
9a + k = 10
a + k = 0
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8a = 10
a = 5/4
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k = -5/4
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y = (5/4)(x-3)^2 - (5/4)
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Cheers,
stan H.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

If the equation of the axis is or alternatively , then the -coordinate of the vertex must be 3.

In the equation:

is the vertex.

Therefore, the value of is given directly and we now have:

If the -intercept is 10, that means the graph includes the point (0,10). In other words:

which can be written:

1:

if one of the -intercepts is 2, then the graph includes the point (2,0). In other words:

which can be written:

2:

Multiply Equation 2 by -1:

2a:

Add Equation 2a to Equation 1:

1a:

Then from Equation 2 we see that

And finally the desired equation is:

Check:

Since is the vertex, is the -coordinate of the vertex. Since the -coordinate of the vertex was determined to be 3, the value of the function, which is to say the value of must be whenever is 3:

: Checks

John