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Find a+b+c if the graph of the equation \(y=ax^2+bx+c\)  is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0).

mathtoo  Mar 12, 2017
 #1
avatar+87334 
+5

Find a+b+c if the graph of the equation   is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0)

 

We have that

 

y = a(x - h)^2 + k  and we know that (h, k)  = (5,3)  so

 

y = a(x - 5)^2 + 3       and since the point (2,0) is on the graph, we have that

 

0 = a (2 - 5)^2 + 3

 

0 = a(-3)^2 + 3

 

0 = 9a + 3

 

-3 = 9a        divide through by 3

 

-1 = 3a

 

-1/3  = a

 

So...our function is

 

y = (-1/3) (x - 5)^2 + 3     simplify

 

y = (-1/3) (x^2 - 10k + 25) + 3

 

y = (-1/3)x^2 + (10/3)x - 25/3 + 3

 

y = (-1/3)x^2 + (10/3)x - 25/3 + 9/3

 

y = (-1/3)x^2 + (10/3)x - 16/3

 

a = (-1/3)  b = (10/3)  and c = (-16/3)    so

 

a + b + c  =   [ -1 + 10 - 16] / 3  =  -7/3

 

Here's the graph : https://www.desmos.com/calculator/chosahxdr6

 

 

cool cool cool

CPhill  Mar 12, 2017
 #2
avatar+7348 
0

Find a+b+c if the graph of the equation is a parabola with vertex (5,3) , vertical axis of symmetry, and contains the point (2,0).

 

\(f(x)=y=ax^2+bx+c\)

 

\(c=3\)

\(x_1=2\)

\(x_2=8\)

\(a \ is \ negative\)

 

\(2 = {-b + \sqrt{b^2-12a} \over 2a}\)                \(8 = {-b- \sqrt{b^2-12a} \over 2a}\)

 

\(16a = {-b - \sqrt{b^2-12a} }\)

.         plus

  \(4a = {-b + \sqrt{b^2-12a} }\)

 

\(20a=-2b\)

 

b=-10a

 

\(4a = {10a - \sqrt{100a^2-12a} }\)

 

\(-6a= - \sqrt{100a^2-12a}\)

 

\(36a^2=100a^2-12a\)

 

\(64a^2=12a\)

 

\(a=\frac{3}{16}\)  Error. CPhill was too fast.

asinus  Mar 12, 2017

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