Question

How do you graph the parabola `y = -2(x -1)^2 + 3` using vertex, intercepts and additional points?

Answer 1

Refer to the explanation.

Explanation:

Graph:

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

This is a quadratic equation in vertex form:

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

where:

`a=-2`, `h=1`, `k=3`

The vertex is the point `(h,k)`, which is `(1,3)`. Plot this point.

The y-intercept is the value of `y` when `x=0`. Substitute `0` for `x` and solve for `y`.

`y=-2(0-1)^2+3`

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

`y=-2(1)+3`

`y=-2+3`

`y=1`

The y-intercept is `(0,1)`. Plot this point.

To find the x-intercepts, convert the equation from vertex form to standard form.

Expand `(x-1)^2` to `x^2-2x+1`.

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

`y=-2x^2+4x-2+3`

`y=-2x^2+4x+1`

The x-intercepts are the values of `x` when `y=0`. Substitute `0` for `y` and solve for `x`.

`0=-2x^2+4x+1`,

where:

`a=-2`, `b=4`, `c=1`

Switch sides.

`-2x^2+4x+1=0`

Solve for the x-intercepts using the quadratic formula.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug in the known values.

`x=(-4+-sqrt(4^2-4*-2*1))/(2*-2)`

Simplify.

`x=(-4+-sqrt(24))/(-4)`

Prime factorize `24`.

`x=(-4+-sqrt(2xx2xx2xx3))/(-4)`

Simplify `2xx2` to `2^2`.

`x=(-4+-sqrt(2^2xx2xx3))/(-4)`

Apply rule `sqrt(a^2)=a`

`x=(-4+-2sqrt6)/(-4)`

`x=(-4+2sqrt6)/(-4)`, `(-4-2sqrt6)/(-4)`

Simplify.

`x=1-(2sqrt6)/2`, `1+(2sqrt6)/2`

The x-intercepts are `(1-(sqrt6)/2,0)` and `(1+(sqrt6)/2,0)`.

The approximate x-intercepts are `(-0.225,0)` and `(2.225,0)`.

Plot the x-intercepts.

Additional point

Choose values for `x` and solve for `y`.

`x=2`

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

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

`y=-2(1)+3`

`y=-2+3`

`y=1`

The additional point is `(2,1)`. Plot this point.

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

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

graph{y=-2(x-1)^2+3 [-10, 10, -5, 5]}