Question 40216: Write the equation for the axis of symmetry of the graph of the equation y=5+16x-2x^2 (Hint: Rewrite the equation so that it fits the standard form of a quadratic function.)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Find the vertex of the graph of the equation y=5+16x-2x^2.
=-2(X^2-2.X*4+4^2)+2*4^2+5
=37-2(X-4)^2
VERTEX IS AT X-4=0....OR...X=4...AT X=4,WE HAVE Y=37..
HENCE VERTEX IS AT (4,37)
Show the algebraic steps used to solve this problem.
Miscellaneous_Word_Problems/40220: The vertex of a parabola is at (-4,-3). If one x-intercept is at (-11,0), what are the coordinates of the other x-intercept?
1 solutions
Answer 25622 by venugopalramana(1956) About Me on 2006-06-01 06:49:22 (Show Source):
The vertex of a parabola is at (-4,-3).
PARABOLA IS SYMMETRIC ABOUT VERTEX..SO X=-4 IS THE LINE OF SYMMETRY
If one x-intercept is at (-11,0),
DISTANCE OF THIS INTERCEPT POINT FROM X=-4 IS -11-(-4)=-7
HENCE THE OTHER INTERCEPTING POINT IS AT -4+7=3
what are the coordinates of the other x-intercept?
COORDINATES OF OTHER INTERCEPT = (3,0)
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