Question

identify the vertex of the quadratic function in vertex form.

1. y = 3(x - 7)^2 - 1
2. y = 3(x + 2)^2 - 5
3. y=(x - 3)^2
4. y=-4(x - 2)^2+4
5. y = 2(x + 1)^2 - 3
6. y=(x + 4)^2
7. y = 1/2(x - 5)^2+1
8. y=-(x + 6)^2+10

Explanation:

Step1: Recall vertex - form of quadratic function

The vertex - form of a quadratic function is \(y = a(x - h)^2+k\), and the vertex is \((h,k)\).

Step2: Find vertex for \(y = 3(x - 7)^2-1\)

For \(y = 3(x - 7)^2-1\), comparing with \(y=a(x - h)^2 + k\), we have \(h = 7\) and \(k=-1\). So the vertex is \((7,-1)\).

Step3: Find vertex for \(y = 3(x + 2)^2-5\)

For \(y = 3(x + 2)^2-5\), rewrite it as \(y=3(x-(- 2))^2-5\). Here \(h=-2\) and \(k = - 5\), so the vertex is \((-2,-5)\).

Step4: Find vertex for \(y=(x - 3)^2\)

For \(y=(x - 3)^2\), we can write it as \(y=(x - 3)^2+0\). So \(h = 3\) and \(k = 0\), and the vertex is \((3,0)\).

Step5: Find vertex for \(y=-4(x - 2)^2+4\)

For \(y=-4(x - 2)^2+4\), we have \(h = 2\) and \(k = 4\), so the vertex is \((2,4)\).

Step6: Find vertex for \(y = 2(x + 1)^2-3\)

Rewrite \(y = 2(x + 1)^2-3\) as \(y=2(x-(-1))^2-3\). Here \(h=-1\) and \(k=-3\), so the vertex is \((-1,-3)\).

Step7: Find vertex for \(y=(x + 4)^2\)

Write \(y=(x + 4)^2\) as \(y=(x-(-4))^2+0\). So \(h=-4\) and \(k = 0\), and the vertex is \((-4,0)\).

Step8: Find vertex for \(y=\frac{1}{2}(x - 5)^2+1\)

For \(y=\frac{1}{2}(x - 5)^2+1\), we have \(h = 5\) and \(k = 1\), so the vertex is \((5,1)\).

Step9: Find vertex for \(y=-(x + 6)^2+10\)

Rewrite \(y=-(x + 6)^2+10\) as \(y=-(x-(-6))^2+10\). Here \(h=-6\) and \(k = 10\), so the vertex is \((-6,10)\).

Answer:

1. \((7,-1)\)
2. \((-2,-5)\)
3. \((3,0)\)
4. \((2,4)\)
5. \((-1,-3)\)
6. \((-4,0)\)
7. \((5,1)\)
8. \((-6,10)\)