To find the equation of the quadratic function represented by the provided points, we can derive it by figuring out the vertex form of the quadratic equation, which is:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola.
Looking at the provided values of \(y\) for different \(x\), it appears that the vertex occurs at the middle of the \(x\) values where \(y\) is maximized. The values of \(y\) are highest at \(x = -2\) and \(x = -1\), suggesting that the vertex might be at:
- The average of \(-2\) and \(-1\) is \(-1.5\), but noticing the symmetry in values around \(-2\), it seems better to select \(-2\) as the vertex.
From the pattern, we can see that:
- The maximum value of \(y\) (which is the value of \(k\)) occurs at \(x = -2\) when \(y = 4\).
- It appears that the parabola opens downwards since \(y\) is decreasing as we move away from \(x = -2\).
To confirm, we can evaluate at \(x = 0\):
Using \(x = 0\): \[ y = a(0 + 2)^2 + 4\] From the table, at \(x = 0\), \(y = 3\): \[ 3 = a(2)^2 + 4 \] \[ 3 = 4a + 4 \] \[ -1 = 4a \] \[ a = -\frac{1}{4} \]
Now, we can fill in our equation in the form \(y = a(x - h)^2 + k\):
Thus, substituting our values:
- \(a = -\frac{1}{4}\),
- \(h = -2\),
- \(k = 4\),
The equation is:
\[ y = -\frac{1}{4}(x + 2)^2 + 4 \]
Now, placing values into the drop-down options:
- Drop down 1: -1/4
- Drop down 2: -2
- Drop down 3: 4
Correct selections are:
y = (-1/4)(x - (-2))^2 + 4.
1 answer