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 ->  Quadratic Equations and Parabolas -> 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      Log On


   

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) About Me  (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)
----
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%285%2F4%29%28x-3%29%5E2-%285%2F4%29%29
----
Cheers,
stan H.

Answer by solver91311(24713) About Me  (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