Question

What are the important points needed to graph `y= 2(x-3)^2+4`?

Answer 1

Refer to the explanation.

Explanation:

Graph:

`y=2(x-3)^2+4` is a quadratic equation in vertex form:

`y=a(x-h)^2+k`,

where:

`a=2`, `h=3`, `k=4`

The important points to graph a parabola are the axis of symmetry, vertex, the y-intercept, x-intercepts (if there are real solutions), and additional points (especially if there are no x-intercepts).

Axis of symmetry: vertical line that divides the parabola into two equal halves. In vertex form, the axis of symmetry is:

`x=h`

`x=3`

Vertex: maximum or minimum point of the parabola

Since `a>0`, the vertex is the minimum point and the parabola opens upward.

The vertex is `(h,k)`, which in this equation is `(3,4)`. Plot this point.

Because the vertex `(3,4)` is above the x-axis, and it is the minimum point, there are no x-intercepts for this equation, so you will need additional points to graph the parabola.

Y-intercept: value of `y` when `x=0`

Substitute `0` for `x` and solve for `y`.

`y=2(0-3)^2+4`

`y=2(-3)^2+4`

`y=2(9)+4`

`y=18+4`

`y=22`

The y-intercept is `(0,22)`. Plot this point.

To determine additional points, choose values for `x` and solve for `y`.

Since the axis of symmetry is `x=3`, you can use this to determine symmetrical points. For example, the y-intercept is `(0,22)` which is three spaces to the left of `3`, so the symmetrical point is three spaces farther than `3`, or `x=6`.

Additional point 1: `x=6`

`y=2(6-3)^2+4`

`y=2(3)^2+4`

`y=2(9)+4`

`y=18+4`

`y=22`

Additional point 1: `(6,22)` Plot this point.

Additional point 2: `x=1`

`y=2(1-3)^2+4`

`y=2(-2)^2+4`

`y=2(4)+4`

`y=12`

Additional point 2: `(1,12)` Plot this point.

Additional point 3: `x=5`

`y=2(5-3)^2+4`

`y=2(2)^2+4`

`y=2(4)+4`

`y=12`

Additional point 3: `(5,12)` Plot this point.

Sketch a parabola through the points. Do not connect the dots.

graph{y=2(x-3)^2+4 [-9.33, 10.67, 22.08, 32.08]}