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

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Answer 1 · 2017-09-12T21:50:44

Vertex: #(-5,5)#

#y#-intercept: #(0,-20)#

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

(See graph at bottom of the explanation)

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]}