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

Sep 12, 2017

Vertex: $\left(- 5 , 5\right)$

$y$-intercept: $\left(0 , - 20\right)$

3rd point, which is a reflection of the $y$-intercept across the axis of symmetry: $\left(- 10 , - 20\right)$

(See graph at bottom of the explanation)

#### Explanation:

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

$- \frac{b}{2 a}$

In the equation you provided, the corresponding values are:

$a = - 1$

$b = - 10$

$c = - 20$

Plugging in the values to find the vertex yields:

$- \frac{- 10}{2 \cdot - 1}$

$= \frac{10}{-} 2$

$= - 5$

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

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

$- {x}^{2} - 10 x - 20$

$= - {\left(- 5\right)}^{2} - 10 \cdot \left(- 5\right) - 20$

$= - 25 + 50 - 20$

$= 5$

Therefore, the vertex is located at the point $\left(- 5 , 5\right)$

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

$- {x}^{2} - 10 x - 20$

$= - {0}^{2} - 10 \cdot 0 - 20$

$= - 20$

Therefore, the $y$-intercept of the vertex is at the point: $\left(0 , - 20\right)$

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: $\left(- 10 , - 20\right)$

Here is the graph. All the points are correct:

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