Hi,  

the vertex form of a parabola,  where(h,k) is the vertex

vertex (1,1)

y =a(x-1)^2 +1

using y-intercept pt(0,3) to find a

3 = a(-1)^2 + 1

2 = a

y = 2(x-1)^2 + 1  OR y = 2[x^2 - 2x +1] + 1

 Vertex Pt(1,1) in the minimum point for the parabola (a>0) 

parabola does not intercept the x-axis



 

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
 The parabola can be defined by the form:

y=A(x-h)^2+k, h,k representing the (x,y) coordinates of the vertex

with the given y-intercept coordinates (0,-3) and vertex coordinates(1,1)

-3=A(0-1)^2+1

A=-4

equation of parabola: y=-4(x-1)^2+1

=-4(X^2-2x+1)+1 =-4x^2+8x-3

to find x-intercept, set y=0

0=-4(x-1)^2+1

4(x-1)^2=1

(x-1)^2=1/4

taking sqrt of both sides,

x-1=±1/2

x=1/2 or 3/2



ans: x -intercepts at 1/2 and 3/2

 

 

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