SOLUTION: The height of an arch is modeled by the equation y=-x^2-6x+16. Which of the following gives the width of the arch at its base? the y-coordinate of the vertex double the x-coordina

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The height of an arch is modeled by the equation y=-x^2-6x+16. Which of the following gives the width of the arch at its base? the y-coordinate of the vertex double the x-coordina      Log On


   

Question 1011493: The height of an arch is modeled by the equation y=-x^2-6x+16. Which of the following gives the width of the arch at its base? the y-coordinate of the vertex double the x-coordinate of the vertex the y-intercept of the equation the difference between the zeroes
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
At the base +y+=+0+, so
+y+=+-x%5E2+-+6x+%2B+16+
+0+=+-x%5E2+-+6x+%2B+16+
+0+=+x%5E2+%2B+6x+-+16+
I can see that:
+0+=+%28+x+%2B+8+%29%2A%28+x+-+2+%29+
+x+=+-8+
+x+=+2+
The difference in these roots is:
+2+-%28-8%29+=+10+
The width of the arch is 10
-------------------------
The center of the arch is +10%2F2+ units
from the roots, so
+2+-+5+=+-3+
and
+-8+%2B+5+=+-3+
so, the x-coordinate of the arch's center is
+x+=+-3+
and
+y+=+-%28-3%29%5E2+-+6%2A%28-3%29+%2B+16+
+y+=+-9+%2B+9+%2B+16+
+y+=+16+
The peak of the arch is at ( -3,16 )