Question 626552: Suppose that f(x)=x^2+4x-12.
A.what is the vertex of f?
B. what are the x-intercepts of the graph of f?
C. solve f(x)= -12 for x. what points are on the graph of f ?
D. use the information obtained in parts A-C to grapgh f(x)=x^2 +4x-12
answers in ordered pair for a through c.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Suppose that f(x)=x^2+4x-12.
complete the square
f(x)=(x^2+4x+4)-12-4
f(x)=(x+2)^2-16
This is an equation of a parabola that opens upwards.
Its standard form: y=(x-h)^2+k (h,k)=(x,y) coordinates of the vertex .
A.what is the vertex of f? (-2,16)
B. what are the x-intercepts of the graph of f? (-6 and 2)
set y=0, then solve for x
0=(x+2)^2-16
(x+2)^2=16
x+2=±√16=±4
x=-6 or x=2
C. solve f(x)= -12 for x. what points are on the graph of f ? (12,180), (-2,16), (0,-6), (0,2)
f(12)=12^2+4*12-12=144+48-12=180
D. use the information obtained in parts A-C to grapgh f(x)=x^2 +4x-12
answers in ordered pair for a through c.
see graph below:
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