Respuesta :
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).
