SOLUTION: 2)For the function y = x2 - 4x - 5 a)Put the function in the form y = a(x - h)2 + k. b)What is the equation for the line of symmetry for the graph of this function? c)Graph the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 2)For the function y = x2 - 4x - 5 a)Put the function in the form y = a(x - h)2 + k. b)What is the equation for the line of symmetry for the graph of this function? c)Graph the      Log On


   

Question 51299: 2)For the function y = x2 - 4x - 5
a)Put the function in the form y = a(x - h)2 + k.
b)What is the equation for the line of symmetry for the graph of this function?
c)Graph the function using the equation in part a. Why it is not necessary to plot points to graph when using y = a (x – h)2 + k.
d)Describe how this graph compares to the graph of y = x2?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
2)For the function y = x2 - 4x - 5
a)Put the function in the form y = a(x - h)2 + k.For the function y = x2 - 4x - 5
={X^2-2(X)(2)+2^2}-2^2-5
=(X-2)^2-9
A=1....H=2.......K=-9
b)What is the equation for the line of symmetry or the graph of this function?
X-2=0....OR...X=2 IS THE LINE OF SYMMETRY
c)Graph the function using the equation in part a. Why it is not necessary to plot points to graph when using y = a (x – h)2 + k.
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-4x-5%2Cx%5E2%29+
DRAW LINE OF SYMMETRY X=2..PLOT VERTEX AT (2,-9)..PLOT CURVE SYMMETRICALLY ALONG THE LINE OF SYMMETRY TAKING THE 2 INTERCEPT POINTS ON THE X AXIS AS
X-2=+3 OR -3....THAT IS X=5 AND -1
d)Describe how this graph compares to the graph of y = x2?
IT IS PARALLEL WITH A SHIFT IN VERTEX FROM (0,0) TO (2,-9)