Question

chapter 3

1. graph: $y = x^2 + 4$

(there is a coordinate grid image here)
factor:

Explanation:

Step1: Identify the function type

The function \( y = x^2 + 4 \) is a quadratic function (parabola) in the form \( y = ax^2 + bx + c \), where \( a = 1 \), \( b = 0 \), \( c = 4 \). Since \( a>0 \), the parabola opens upward.

Step2: Find the vertex

For a parabola \( y = ax^2 + bx + c \), the x - coordinate of the vertex is \( x=-\frac{b}{2a} \). Here, \( b = 0 \), \( a = 1 \), so \( x = 0 \). Substitute \( x = 0 \) into the function: \( y=0^2 + 4=4 \). So the vertex is at \( (0,4) \).

Step3: Find additional points

• When \( x = 1 \): \( y=1^2 + 4=5 \), so the point is \( (1,5) \).
• When \( x=- 1 \): \( y=(-1)^2 + 4=5 \), so the point is \( (-1,5) \).
• When \( x = 2 \): \( y=2^2+4 = 8 \), so the point is \( (2,8) \).
• When \( x=-2 \): \( y=(-2)^2 + 4=8 \), so the point is \( (-2,8) \).

Step4: Plot the points and draw the parabola

Plot the vertex \( (0,4) \), and the points \( (1,5) \), \( (-1,5) \), \( (2,8) \), \( (-2,8) \) on the coordinate plane. Then draw a smooth upward - opening parabola passing through these points.

Answer:

To graph \( y=x^{2}+4 \):

1. Recognize it's an upward - opening parabola (since \( a = 1>0 \)) with vertex at \( (0,4) \).
2. Plot the vertex \( (0,4) \) and additional points like \( (1,5) \), \( (-1,5) \), \( (2,8) \), \( (-2,8) \).
3. Draw a smooth curve through these points to form the parabola.