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Solution
If the x-intercepts are (−3−√2) and (−3+√2) then the x value of the vertex is halfway between these value. So, we add them together and divide by 2. (−3−√2) + (−3+√2) = -6 and -6/2 = -3 so, the x coordinate of the vertex is -3.
The generic equation for a parabola is y = a(x – h)2 + k where h is the x coordinate of the vertex and k is the y coordinate of the vertex. So, we can plug in the -3 as the x coordinate of the vertex making the equation:
y = a(x – (-3))2 + k or y = a(x + 3)2 + k
But we still have two missing values, “a” and “k”. To find them, plug in the y-intercept (0, -5) as the x and y to get:
-5 = a(0 + 3)2 + k which, simplified, gives us -5 = 9a + k
Now plug in one of the x-intercepts, how about (−3−√2, 0) which gives us: 0 = a(−3−√2 + 3)2 + k or
0 = a(−√2)2 + k or 0 = 2a + k
Now we have two equations with two unknowns:
9a + k = -5
2a + k = 0 we can subtract equation 2 from equation 1:
————-
7a = -5
a = -5/7
Then plug -5/7 in for a to find k:
2(-5/7) + k = 0 or k = 10/7
So the equation is y = -5/7(x + 3)2 + 10/7
This could be multiplied out to get the equation in standard form. -5/7(x + 3)2 + 10/7= -5/7(x2 + 6x + 9) + 10/7 = -5/7×2 – 30/7x – 45/7 + 10/7 or y = -5/7×2 – 30/7x – 35/7 or y = -5/7×2 – 30/7x – 5