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
Algebra.Com
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
----
You have three points: (0,10), (2,0) and vertex(3,y)
---
If vertex is (3,y), h=3.
y = a(x-3)^2+k
-----
Substituting (0,10) and (2,0)
10 = a(-3)^2 + k
0 = a(-1)^2 + k
---
Solve for a and k:
9a + k = 10
a + k = 0
----------------------
8a = 10
a = 5/4
---
k = -5/4
--------------------
y = (5/4)(x-3)^2 - (5/4)
----
----
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:
^2\ +\ k)
)
is the vertex.
Therefore, the value of 
is given directly and we now have:
^2\ +\ k)
If the 
-intercept is 10, that means the graph includes the point (0,10). In other words:
^2\ +\ k)
which can be written:
1: 
if one of the -intercepts is 2, then the graph includes the point (2,0). In other words:
^2\ +\ k)
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:
^2\ -\ \frac{10}{8})
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:
^2\ -\ \frac{10}{8})
: Checks
John
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