Write the formula for the parabola that has x- intercepts (−3−sqrt(2) ,0) and (−3+sqrt(2) ,0), and y- intercept (0,−5)

Answer 1

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:

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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 = -5/7x² - 30/7x - 45/7 + 10/7 or y = -5/7x² - 30/7x - 35/7 or y = -5/7x² - 30/7x - 5