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avatar+388 

The parabola $y = ax^2 + bx + c$ is graphed below. Find $a \cdot b \cdot c.$ (The grid lines are one unit apart.)

 

 Jun 6, 2024
 #1
avatar+759 
+1

First, let's write this in vertex form. 

We have \(y = a(x-h)^2 + k\) where \((h,k) \) denotes the vertex. From the graph, we can write, \(y=a(x-3)^2+1\) since the vertex is at (3, 1)

 

Now, we have to find a. We can use the y intercept to help. We have 

\(7 = a (0 - 3)^2 + 1 \\ 6 = 9a \\ a = 6/9 = 2/3\)

 

We find that when \(y=2/3(x-3)^2+1\), we have a y intercept at (0, 7), meaning we found our equation. 

 

Now, we simply have to expand it. Expaning, we get \(y = \frac{2}{3} x^ 2 − 4 x + 7\)

 

This means \(abc = 2/3(-4)7 = -56/3\)

 

-56/3 is our answer!

 

Thanks! :)

 Jun 6, 2024
edited by NotThatSmart  Jun 6, 2024
 #2
avatar+129401 
+1

We have the vertex form

 

y = a(x - h)^2  + k         where the vertex  (h,k)  = (3,1)

 

We  know that  (0,7)  is on the graph..so....

 

7  = a (0 - 3)^2  + 1

 

6 =  9a

 

a = 6/9  = 2/3

 

So

 

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

 

y = (2/3) (x^2 - 6x + 9) + 1

 

y = (2/3)x^2 - 4x +6 + 1

 

y = (2/3)x^2 - 4x + 7

 

 

a*b*c =  (2/3) (-4) (7)  =  -56 / 3

 

cool cool cool

 Jun 6, 2024

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