Questions: Graphing a parabola of the form y = ax² + c Graph the parabola: y = (1/2)x^2 - 1 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. [Grid and graphing interface]

Graphing a parabola of the form y = ax² + c

Graph the parabola:

y = (1/2)x^2 - 1

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.

[Grid and graphing interface]

Transcript text: www.awu.aleks.com Graphing a parabola of the form y = ax² + c Graph the parabola: $y = \frac{1}{2}x^2 - 1$ 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. [Grid and graphing interface] Explanation Check

Solution

Solution Steps

Step 1: Find the vertex

The equation is in the form $y = ax^2 + c$. The vertex of a parabola in this form is at $(0, c)$. In this case, $c = -3$, so the vertex is $(0, -3)$.

Step 2: Choose x-values

We need two x-values to the left and two to the right of the vertex's x-value (0). We'll choose -2, -1, 1, and 2.

Step 3: Calculate y-values

Substitute each x-value into the equation $y = \frac{1}{2}x^2 - 3$:

$x = -2$: $y = \frac{1}{2}(-2)^2 - 3 = \frac{1}{2}(4) - 3 = 2 - 3 = -1$
$x = -1$: $y = \frac{1}{2}(-1)^2 - 3 = \frac{1}{2}(1) - 3 = \frac{1}{2} - 3 = -\frac{5}{2} = -2.5$
$x = 1$: $y = \frac{1}{2}(1)^2 - 3 = \frac{1}{2}(1) - 3 = \frac{1}{2} - 3 = -\frac{5}{2} = -2.5$
$x = 2$: $y = \frac{1}{2}(2)^2 - 3 = \frac{1}{2}(4) - 3 = 2 - 3 = -1$