+595 The parabola $y = ax^2 + bx + c$ is graphed below. Find $a \cdot b \cdot c.$ (The grid lines are one unit apart.)
+1894 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! :)
+129847 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
