Question

How to graph a parabola `y=-3(x-2)^2 + 5`?

Answer 1

Refer to the explanation.

Explanation:

Graph:

`y=-3(x-2)^2+5` is a quadratic equation in vertex form:

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

where:

`h` is the axis of symmetry and `(h,k)` is the vertex.

In order to graph a parabola, you need the vertex, the y-intercept, x-intercepts, and one or more additional points.

Vertex: maximum or minimum point of the parabola. Since `a<0`, the vertex is the maximum point and the parabola opens downward.

The vertex is `(2,5)`

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

Substitute `0` for `x`.

`y=-3(0-2)^2+5`

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

`y=-3(4)+5`

`y=-7`

The y-intercept is `(0,-7)`.

X-intercepts (Roots): values for `x` when `y=0`

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

`0=-3(x-2)^2+5`

Subtract `5` from both sides.

`-5=-3(x-2)^2`

Divide both sides by `-3`.

`5/3=(x-2)^2` `larr` Two negatives make a positive.

Take the square root of both sides.

`+-sqrt(-5/3)=x-2`

Use the rule `sqrt(a/b)=(sqrta)/(sqrtb)`

`+-(sqrt5)/(sqrt3)=x-2`

Add `2` to both sides.

`2+-(sqrt5)/(sqrt3)=x`

Switch sides.

`x=2+-(sqrt5)/(sqrt3)`

`x=2+(sqrt5)/(sqrt3)`, `2-(sqrt5)/(sqrt3)`

`x=~~3.291, 0.709`

The x-intercepts are: `(2+(sqrt5)/(sqrt3),0)`, `(2-(sqrt5)/(sqrt3))`

Approximate x-intercepts: `(~~3.291,0)`, `(0.709,0)`

Additional point:

Substitute `4` for `x` and solve for `y`.

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

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

`y=-3(4)+5`

`y=-12+5`

`y=-7`

Additional point: `(4,-7)`

Summary:

Vertex: `(2,5)`

Y-intercept: `(0,-7)`

X-intercepts: `(2+(sqrt5)/(sqrt3),0)`, `(2-(sqrt5)/(sqrt3))`

Approximate x-intercepts: `(~~3.291,0)`, `(0.709,0)`

Additional point: `(4,-7)`

Plot the points and sketch a parabola through the points. Do not connect the dots.

graph{y=-3(x-2)^2+5 [-10, 10, -5, 5]}