Find the x-intercepts of the parabola withvertex (1,1) and y-intercept (0,-3).Write your answer in this form: (X1,y,),(X2,).If necessary, round to the nearest hundredth.​

Answer:[tex](\frac{1}{2},0),(\frac{3}{2},0)[/tex]Step-by-step explanation:The vertex form of a parabola is given by:[tex]y=a(x-h)^2+k[/tex], where V(h,k) is the vertex of the parabola.The given parabola has vertex (1,1).This implies that: [tex]h=1,k=1[/tex].Put these values into the vertex form equation.[tex]\implies y=a(x-1)^2+1[/tex]The y-intercept of this parabola is: (0,-3).This point lies on the parabola hence it must satisfy its equation.[tex]\implies -3=a(0-1)^2+1[/tex][tex]\implies -3=a(-1)^2+1[/tex][tex]\implies -3=a(1)+1[/tex][tex]\implies -3=a+1[/tex][tex]\implies -3-1=a[/tex][tex]\implies -4=a[/tex]The equation now becomes [tex]\implies y=-4(x-1)^2+1[/tex]To find the x-intercept, put y=0 into the equation:[tex]\implies -4(x-1)^2+1=0[/tex][tex]\implies -4(x-1)^2=-1[/tex]Divide through by -4.[tex]\implies \frac{-4(x-1)^2}{-4}=\frac{-1}{-4}[/tex][tex]\implies (x-1)^2=\frac{-1}{-4}[/tex][tex]\implies (x-1)^2=\frac{1}{4}[/tex]Take plus or minus square root of both sides.[tex]\implies x-1=\pm \sqrt{\frac{1}{4}}[/tex][tex]\implies x-1=\pm \frac{1}{2}[/tex][tex]\implies x=1\pm \frac{1}{2}[/tex][tex]\implies x=1-\frac{1}{2}[/tex] or [tex]\implies x=1+ \frac{1}{2}[/tex][tex]\implies x=\frac{1}{2}[/tex] or [tex]\implies x=1 \frac{1}{2}[/tex]Therefore the x-intercepts are:[tex](\frac{1}{2},0),(\frac{3}{2},0)[/tex]To the nearest hundredth, we have [tex]0.50,0),(1.50,0)[/tex]