SOLUTION: Find the equation of the quadratic function satisfying the given conditions. ​(​Hint: Determine values of​ a, h, and k that satisfy ​P(x)=​a(x-​h)^2+​k.) Express the

Algebra ->  Equations -> SOLUTION: Find the equation of the quadratic function satisfying the given conditions. ​(​Hint: Determine values of​ a, h, and k that satisfy ​P(x)=​a(x-​h)^2+​k.) Express the       Log On


   

Question 1191949: Find the equation of the quadratic function satisfying the given conditions. ​(​Hint: Determine values of​ a, h, and k that satisfy ​P(x)=​a(x-​h)^2+​k.) Express the answer in the form ​P(x)=ax^2+bx+c. Use a calculator to support the result.
Vertex: (-3,-5); through (5,123)
P(x)=
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
y=ax^2+bx+c
-5=9a-3b+c
123=25a+5b+c
subtract 2 from 1
-128=-16a-8b
or
2a+b=16, multiplying by -1, dividing by 8, and rearranging the terms.
also
y=a(x+3)^2-5
123=64a-5
64a=128
a=2
b=12
y=2x^2+12x+13, the 13 comes from 2(x+3)^2-5. The constant is 2*3^2-5
-
-5=9a-3b+c, a=2 b=12 c=13 checks
123=25a+5b+c same
P(x)=2x^2+12x+13,
graph%28300%2C300%2C-10%2C10%2C-15%2C15%2C2x%5E2%2B12x%2B13%29
graph%28300%2C300%2C-10%2C20%2C-25%2C200%2C2x%5E2%2B12x%2B13%29

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The answer is found more easily if you use the given hint.

The vertex form of the equation is

P%28x%29=a%28x-h%29%5E2%2Bk

where (h,k) is the vertex and a is a constant that determines the "steepness" of the parabola. The vertex is given...

P%28x%29=a%28x%2B3%29%5E2-5

... and the constant a is quickly determined using the other given point.

123=a%288%5E2%29-5
128=64a
a=2

The vertex form is P%28x%29=2%28x%2B3%29%5E2-5

Then

2%28x%2B3%29%5E2-5+=+2%28x%5E2%2B6x%2B9%29-5+=+2x%5E2%2B12x%2B18-5+=+2x%5E2%2B12x%2B13

ANSWER: P(x) = 2x^2+12x+13