A parabola with vertex (1,5) and y-intercept(0,2) crosses the x-axis in two places. One x-intercept is at (-0.29,0). Find the other x-intercept. Separate the values with a comma.

Answer:So the other x-intercept we are looking for is (2.29 , 0).Step-by-step explanation:The equation for a parabola in vertex form is [tex]y=a(x-h)^2+k[/tex] where (h,k) is the vertex.So we are given (h,k)=(1,5) so let's plug that in.  This gives us the following equation for our parabola:[tex]y=a(x-1)^2+5[/tex].Now we need to find [tex]a[/tex]. Let's find [tex]a[/tex] by using another point (x,y) given.  We are given that (0,2) is on our parabola. So when x is 0, y is 2.This gives us the equation:[tex]2=a(0-1)^2+5[/tex][tex]2=a(-1)^2+5[/tex][tex]2=a(1)+5[/tex][tex]2=a+5[/tex][tex]2-5=a[/tex][tex]-3=a[/tex]So our parabola in vertex form looks like this:[tex]y=-3(x-1)^2+5[/tex]Now we are asked to find the x-intercepts.You can find the x-intercepts by setting y equal to 0 and solving for x.So let's do that:[tex]0=-3(x-1)^2+5[/tex]Subtract 5 on both sides:[tex]-5=-3(x-1)^2[/tex]Divide both sides by -3:[tex]\frac{5}{3}=(x-1)^2[/tex]Take the square root of both sides:[tex]\pm \sqrt{\frac{5}{3}}=x-1[/tex]Add 1 on both sides:[tex]\pm \sqrt{\frac{5}{3}}+1=x[/tex]So the two solutions in exact form are[tex]x=\sqrt{\frac{5}{3}}+1 \text{ or } -\sqrt{\frac{5}{3}}+1[/tex]Putting both into calculator (separately) gives:[tex]x \approx 2.29 \text{ or } -0.29[/tex]So the other x-intercept we are looking for is (2.29 , 0).