The other x intercept would be 2.29.
You can find this by first modeling the equation using vertex form: y = a(x - h)^2 + k, in which the variables stand for the following:
a = constant
h = x value of vertex
k = y value of vertex.
Since we know enough to plug in and solve for a, we can get the equation we are using.
y = a(x - h)^2 + k
First put in the vertex points into (h, k)
y = a(x - 1)^2 + 5
Now add the points we know for x and y from the y-intercept
2 = a(0 - 1)^2 + 5
And solve that for a
2 = a(0 - 1)^2 + 5
2 = a (-1)^2 + 5
2 = a + 5
-3 = a
Now we can model the equation as y = -3(x - 1)^2 + 5. Using that, we can distribute to put in standard form.
y = -3(x - 1)^2 + 5
y = -3(x^2 - 2x + 1) + 5
y = -3x^2 + 6x - 3 + 5
y = -3x^2 + 6x + 2
Now we can use the quadratic equation to solve for the remaining zero. Using the following variables:
a = -3 (attached to x^2)
b = 6 (attached to x)
c = 2 (constant)
![\frac{-b +/- \sqrt{b^(2) - 4ac}}{2a}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3cvjkrjyc8poioffdj1eccd706lbnug31r.png)
![\frac{-6 +/- \sqrt{6^(2) - 4(-3)(2)}}{2(-3)}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/5rqcroaqgl81xbcrutyinydhzkc2rm6k6y.png)
![(-6 +/- √(36 + 24))/(-6)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/kxeatbdztsscbwoq3bwsu6karj1mnrxxim.png)
![(-6 +/- √(60))/(-6)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/uadacg7ft8dleabctaoup2sw3lz2tnwyjy.png)
![(-6 +/- 7.74)/(-6)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/58w3n8dsv49odh1vtfmyun8byqc10sra8q.png)
x = 0.29 or x = -2.29