How do you graph the parabola `y=-x^2-10x-20` using vertex, intercepts and additional points?

To find the `x`-coordinate of the vertex of a parabola from a quadratic equation in standard form, use the formula:

`-b/(2a)`

In the equation you provided, the corresponding values are:

`a=-1`

`b=-10`

`c=-20`

Plugging in the values to find the vertex yields:

`-(-10)/(2*-1)`

`=10/-2`

`=-5`

That's the `x`-coordinate of the vertex.

To find the `y`-coordinate, plug in the `x`-coordinate value into the original quadratic equation:

`-x^2-10x-20`

`= -(-5)^2-10*(-5)-20`

`= -25+50-20`

`= 5`

Therefore, the vertex is located at the point `(-5,5)`

To find the `y`-intercept, just make the value of `x=0`:

`-x^2-10x-20`

`=-0^2-10*0-20`

`=-20`

Therefore, the `y`-intercept of the vertex is at the point: `(0,-20)`

To find another point, simply draw a vertical line that intersects the vertex. That's called a line of symmetry.

Since you have 2 points, look at the `y`-intercept and imagine a point reflected across the line of symmetry.

That point will be your last point, and is located at: `(-10,-20)`

Here is the graph. All the points are correct:

graph{-x^2-10x-20 [-28.97, 22.35, -20.29, 5.35]}