Question

1. graph each quadratic function $y = -3x^2 + 2$.

Explanation:

Step1: Identify the vertex form

The quadratic function is given in the form \( y = ax^2 + k \), which is a vertical shift of the parent function \( y = x^2 \). Here, \( a=-3 \) and \( k = 2 \). The vertex of \( y=ax^2 + k \) is at \( (0,k) \), so the vertex is \( (0,2) \).

Step2: Determine the direction of opening

Since \( a=-3<0 \), the parabola opens downward.

Step3: Find additional points

• When \( x = 1 \): \( y=-3(1)^2+2=-3 + 2=-1 \), so the point is \( (1,-1) \).
• When \( x=-1 \): \( y=-3(-1)^2+2=-3 + 2=-1 \), so the point is \( (-1,-1) \).
• When \( x = 2 \): \( y=-3(2)^2+2=-12 + 2=-10 \), so the point is \( (2,-10) \).
• When \( x=-2 \): \( y=-3(-2)^2+2=-12 + 2=-10 \), so the point is \( (-2,-10) \).

Step4: Plot the vertex and points

Plot the vertex \( (0,2) \), then the points \( (1,-1) \), \( (-1,-1) \), \( (2,-10) \), \( (-2,-10) \) (or other symmetric points) on the coordinate grid. Then draw a smooth, downward - opening parabola through these points.

Answer:

To graph \( y=-3x^{2}+2 \):

1. Plot the vertex at \( (0,2) \).
2. Plot symmetric points like \( (1, - 1) \), \( (-1, - 1) \), \( (2, - 10) \), \( (-2, - 10) \) (or others).
3. Draw a smooth, downward - opening parabola through these points.