Question

3 graph the parabola from the function given below.

(g(x) = 5(x + 1)^2 - 7)

Explanation:

🆕 New Concept Discovered: Graphing Quadratic Functions in Vertex Form
Finding the turning point and key points of a parabola

Step 1: Identify the vertex

The given quadratic function is in vertex form:

$$ g(x) = a(x - h)^2 + k $$

Comparing this to our function:

$$ g(x) = 5(x + 1)^2 - 7 $$

We can rewrite \( (x + 1) \) as \( (x - (-1)) \):

• \( h = -1 \)
• \( k = -7 \)

The vertex \( (h, k) \) is the lowest point (turning point) of the parabola:

$$ \text{Vertex} = (-1, -7) $$

Step 2: Find additional points

To graph the parabola accurately, we calculate the function values for \( x \)-values near the vertex.

For \( x = 0 \):

$$ g(0) = 5(0 + 1)^2 - 7 = 5(1)^2 - 7 = 5 - 7 = -2 $$

This gives the point \( (0, -2) \).

For \( x = -2 \) (by symmetry about the axis of symmetry \( x = -1 \)):

$$ g(-2) = 5(-2 + 1)^2 - 7 = 5(-1)^2 - 7 = 5 - 7 = -2 $$

This gives the point \( (-2, -2) \).

Step 3: Plot the points and draw the parabola

1. Select the PARABOLA tool.
2. Click on the vertex at \( (-1, -7) \).
3. Click on one of the other calculated points, such as \( (0, -2) \) or \( (-2, -2) \), to define the width and direction of the parabola.

Answer:

To graph the function \( g(x) = 5(x + 1)^2 - 7 \):

1. Plot the vertex at \( (-1, -7) \).
2. Plot a second point at \( (0, -2) \) (or \( (-2, -2) \)).
3. Draw the upward-opening parabola passing through these points.