Find the standard equation of a parabola with zeros at x=2 and x=15 passing through the point (6,−5).
Remember to write your answers as exact numbers (fractions), rather than a decimal approximation.
thank you
(x-2)(x-15) = x^2 - 17 x + 30 = 4a(y-k)
if x = 6, y = -5 and also if x = (2+15)/2 , y = 0
36 - 102 + 30 = 4 a (-5-k)
-36 = -20 a -4 k a
and
(17/2)^2 - 17(17/2) + 30 = -4ak
so
-36 = -20 a - 17^2/4 + 30
20 a = 66 + 17^2/4
20 a = (264 + 289)/4 = 553/4
a = 553 / 80
plug and chug
How do write that in standard dorm
y=...(x+-)^2+-.....
help anonymous
thank you
To find the standard equation of a parabola with zeros at x=2 and x=15 passing through the point (6, -5), we can use the fact that if a quadratic function has zeros at x=a and x=b, the standard equation of the parabola can be written as:
f(x) = a(x - a)(x - b)
In this case, the zeros are x=2 and x=15, so our equation would look like:
f(x) = a(x - 2)(x - 15)
To determine the value of the constant a, we can substitute the coordinates of the given point (6, -5) into the equation and solve for a.
Substituting x = 6 and y = -5, we get:
-5 = a(6 - 2)(6 - 15)
Simplifying:
-5 = a(4)(-9)
-5 = -36a
Now, we can solve for a by dividing both sides of the equation by -36:
a = (-5) / (-36) = 5/36
So, the equation of the parabola is:
f(x) = (5/36)(x - 2)(x - 15)
This is the standard form of the equation of the parabola with zeros at x=2 and x=15 passing through the point (6, -5).