Practice 5-3: Write the equation of the parabola in vertex form.

To solve the problem of writing the equation of each parabola in vertex form, we need to follow these steps:

Vertex Form of a Parabola

The vertex form of a parabola is given by:
\[
y = a(x - h)^2 + k
\]
where:
- \((h, k)\) is the vertex of the parabola,
- \(a\) determines the direction and the width of the parabola:
- If \(a > 0\), the parabola opens upwards.
- If \(a < 0\), the parabola opens downwards.
- The absolute value of \(a\) affects the width: larger \(|a|\) makes the parabola narrower, and smaller \(|a|\) makes it wider.

Steps to Solve Each Problem

1. Identify the vertex \((h, k)\) of the parabola from the graph.
2. Determine whether the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)).
3. Use a point on the parabola (other than the vertex) to find the value of \(a\).
4. Write the equation in vertex form.

Let's solve each problem step by step.

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Problem 1

#### Graph:
- Vertex: \((0, 0)\)
- The parabola opens upwards, so \(a > 0\).
- A point on the parabola: \((2, 4)\).

#### Solution:
1. Vertex: \((h, k) = (0, 0)\)
2. Equation: \(y = a(x - 0)^2 + 0 \Rightarrow y = ax^2\)
3. Use the point \((2, 4)\):
\[
4 = a(2)^2 \Rightarrow 4 = 4a \Rightarrow a = 1
\]
4. Equation: \(y = x^2\)

Answer for Problem 1:
\[
\boxed{y = x^2}
\]

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Problem 2

#### Graph:
- Vertex: \((0, 3)\)
- The parabola opens upwards, so \(a > 0\).
- A point on the parabola: \((2, 6)\).

#### Solution:
1. Vertex: \((h, k) = (0, 3)\)
2. Equation: \(y = a(x - 0)^2 + 3 \Rightarrow y = ax^2 + 3\)
3. Use the point \((2, 6)\):
\[
6 = a(2)^2 + 3 \Rightarrow 6 = 4a + 3 \Rightarrow 4a = 3 \Rightarrow a = \frac{3}{4}
\]
4. Equation: \(y = \frac{3}{4}x^2 + 3\)

Answer for Problem 2:
\[
\boxed{y = \frac{3}{4}x^2 + 3}
\]

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Problem 3

#### Graph:
- Vertex: \((1, 0)\)
- The parabola opens upwards, so \(a > 0\).
- A point on the parabola: \((2, 1)\).

#### Solution:
1. Vertex: \((h, k) = (1, 0)\)
2. Equation: \(y = a(x - 1)^2 + 0 \Rightarrow y = a(x - 1)^2\)
3. Use the point \((2, 1)\):
\[
1 = a(2 - 1)^2 \Rightarrow 1 = a(1)^2 \Rightarrow a = 1
\]
4. Equation: \(y = (x - 1)^2\)

Answer for Problem 3:
\[
\boxed{y = (x - 1)^2}
\]

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Problem 4

#### Graph:
- Vertex: \((0, 0)\)
- The parabola opens downwards, so \(a < 0\).
- A point on the parabola: \((2, -4)\).

#### Solution:
1. Vertex: \((h, k) = (0, 0)\)
2. Equation: \(y = a(x - 0)^2 + 0 \Rightarrow y = ax^2\)
3. Use the point \((2, -4)\):
\[
-4 = a(2)^2 \Rightarrow -4 = 4a \Rightarrow a = -1
\]
4. Equation: \(y = -x^2\)

Answer for Problem 4:
\[
\boxed{y = -x^2}
\]

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Problem 5

#### Graph:
- Vertex: \((2, -1)\)
- The parabola opens upwards, so \(a > 0\).
- A point on the parabola: \((4, 3)\).

#### Solution:
1. Vertex: \((h, k) = (2, -1)\)
2. Equation: \(y = a(x - 2)^2 - 1\)
3. Use the point \((4, 3)\):
\[
3 = a(4 - 2)^2 - 1 \Rightarrow 3 = a(2)^2 - 1 \Rightarrow 3 = 4a - 1 \Rightarrow 4a = 4 \Rightarrow a = 1
\]
4. Equation: \(y = (x - 2)^2 - 1\)

Answer for Problem 5:
\[
\boxed{y = (x - 2)^2 - 1}
\]

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Problem 6

#### Graph:
- Vertex: \((-2, -3)\)
- The parabola opens downwards, so \(a < 0\).
- A point on the parabola: \((0, -7)\).

#### Solution:
1. Vertex: \((h, k) = (-2, -3)\)
2. Equation: \(y = a(x + 2)^2 - 3\)
3. Use the point \((0, -7)\):
\[
-7 = a(0 + 2)^2 - 3 \Rightarrow -7 = a(2)^2 - 3 \Rightarrow -7 = 4a - 3 \Rightarrow 4a = -4 \Rightarrow a = -1
\]
4. Equation: \(y = -(x + 2)^2 - 3\)

Answer for Problem 6:
\[
\boxed{y = -(x + 2)^2 - 3}
\]

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Final Answers:

1. \(\boxed{y = x^2}\)
2. \(\boxed{y = \frac{3}{4}x^2 + 3}\)
3. \(\boxed{y = (x - 1)^2}\)
4. \(\boxed{y = -x^2}\)
5. \(\boxed{y = (x - 2)^2 - 1}\)
6. \(\boxed{y = -(x + 2)^2 - 3}\)