How do you graph the parabola #y= - x^2 - 6x - 8# using vertex, intercepts and additional points?
2 Answers
See below
Explanation:
Firstly, complete the square to put the equation in vertex form,
This implies that the vertex, or local maximum (since this is a negative quadratic) is
The quadratic can also be factorised,
which tells us that the quadratic has roots of -2 and -4, and crosses the
Finally, we observe that if we plug
All of this gives us enough information to sketch the curve:
graph{-x^2-6x-8 [-10, 10, -5, 5]}
First, turn this equation to vertex form:
So the
To find the
The
You can also use the quadratic formula to solve if it is not factorable (A discriminant that is a perfect square indicates that the equation is factorable):
The
The y-intercept here is
To find additional points, plug in values for
etc.
A graph below is for reference:
graph{-x^2-6x-8 [-12.295, 7.705, -7.76, 2.24]}