Question

graph the parabola.
$y = \frac{1}{2}x^2 - 3$
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button.

Explanation:

Step1: Find the vertex

The equation of the parabola is in the form \( y = ax^2 + k \), where the vertex is at \( (0, k) \). For \( y=\frac{1}{2}x^2 - 3 \), the vertex is \( (0, - 3) \).

Step2: Find points to the left of the vertex (x negative)

• Let \( x=-2 \): \( y=\frac{1}{2}(-2)^2 - 3=\frac{1}{2}(4)-3 = 2 - 3=-1 \), so the point is \( (-2, -1) \).
• Let \( x = - 4 \): \( y=\frac{1}{2}(-4)^2-3=\frac{1}{2}(16)-3 = 8 - 3 = 5 \), so the point is \( (-4,5) \).

Step3: Find points to the right of the vertex (x positive)

• Let \( x = 2 \): \( y=\frac{1}{2}(2)^2-3=\frac{1}{2}(4)-3 = 2 - 3=-1 \), so the point is \( (2, -1) \).
• Let \( x = 4 \): \( y=\frac{1}{2}(4)^2-3=\frac{1}{2}(16)-3 = 8 - 3 = 5 \), so the point is \( (4,5) \).

Answer:

Vertex: \((0, -3)\), Left points: \((-2, -1)\), \((-4, 5)\), Right points: \((2, -1)\), \((4, 5)\) (These points can be plotted on the graph to draw the parabola)