Question

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

Answer 1

Give the quadratic equation

$y=x^2+2$

To determine the coordinates of the vertex, the first step is to calculate the x-coordinate using the formula:

$x=-\frac{b}{2a}$

Where

a is the coefficient of the quadratic term

b is the coefficient of the x-term

For the given quadratic equation, the coefficient of the quadratic term is a=1, and, since there is no x-term, the value of b is equal to zero, then the x-coordinate of the vertex can be calculated as follows:

$\begin{gathered} x=-\frac{0}{2\cdot1} \\ x=-\frac{0}{2} \\ x=0 \end{gathered}$

Replace the value on the formula of the equation to determine the y-coordinate of the vertex:

$\begin{gathered} y=x^2+2 \\ y=0^2+2 \\ y=2 \end{gathered}$

The coordinates of the vertex are (0,2)

Next, you have to determine the coordinates for 4 points of the parabola, 2 with negative x-coordinate and 2 with positive x-coordinate.

For example for the values x=-3, x=-2, x=2, and x=3

For x=-3

$\begin{gathered} y=(-3)^2+2 \\ y=9+2 \\ y=11 \end{gathered}$

The ordered pair is (-3,11)

For x=-2

$\begin{gathered} y=(-2)^2+2 \\ y=4+2 \\ y=6 \end{gathered}$

The ordered pair is (-2,6)

For x=2

$\begin{gathered} y=2^2+2 \\ y=4+2 \\ y=6 \end{gathered}$

The ordered pair is (2,6)

For x=3

$\begin{gathered} y=3^3+2 \\ y=9+2 \\ y=11 \end{gathered}$

The ordered pair is (3,11)

Plot the five points and draw the curve: