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=-(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=-(0)/(2\cdot1) \\ x=-(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: