Question 834968: 2 parabolas have the same x intercepts (-2,0) and (4,0). The maximum or minimum value of the first is 2 times the maximum or minimum value of the other. how do I find the vetex location for each of these?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if your roots of the equation are:
x = -2 and x = 4, then the factors of the equation must be:
(x+2) * (x-4) = 0
multiply these factors together and you get one form of the equation that has those roots.
that equation is:
x^2 - 2x - 8 = 0
you can graph this equation by setting y = x^2 - 2x - 8.
we won't do that yet, but we will further down after we find the other equation.
i'm going to assume this is the second equation, because then i only have to multiply by 2 rather than divide by 2 which is messier.
so the second equation is:
y = x^2 - 2x - 8
the x-value of the vertex of this equation is the line of symmetry of this equation and is bound by the following formula:
x = -b/2a
a is the coefficient of the x^2 term.
b is the coefficient of the x term.
x = -b/2a becomes x = 2/2 which becomes x = 1
the y-value of the vertex of this equation is found by finding f(-b/2a) which becomes f(1).
this is found by replacing x with 1 in the equation to get:
x^2 - 2x - 8 = 1^2 - 2*1 - 8 which is equal to 1 - 2 - 8 which is equal to -9.
the vertex of the second equation is (1,-9).
since the coefficient of the quadratic equation is positive, this will be a minimum point.
now to the first equation.
the minimum point of the first equation is 2 times the minimum point of the second equation.
this makes the minimum point of the first equation equal to (1,-18)
the x-value is the same.
the y-value is 2 times the y-value of the second equation.
instead of having f(1) = -9, we now have f(1) is equal to -18.
this means that 2 * f(1) is equal to -18.
since 2 * f(1) is equal to 2 * f(x) when x is equal to 1, then it appears that our first equation must be equal to 2 * f(x) which is 2 * (x^2 - 2x - 8) which makes our first equation equal to 2x^2 - 4x - 16.
if we did this correctly, the first equation must have the same x intercepts and must also have a minimum value double that of the second equation.
let's see if we can get that to happen.
we start with:
f(x) = 2x^2 - 4x - 16
we set it equal to 0 to get:
2x^2 -4x - 16 = 0
we factor out the greatest common factor to get:
2 * (x^2 - 2x - 8) = 0
we factor to get:
2 * (x-4) * (x+2) = 0
our roots are the same as the second equation.
our minimum value needs to be calculated off of the original equation which is:
2x^2 - 4x - 16.
a is equal to 2
b is equal to -4
x = -b/2a becomes x = 4/4 which is equal to 1.
we have the same x-value.
the y-value of the minimum point is f(1) which is equal to:
2*1^2 - 4*1 - 16 which is equal to:
2 - 4 - 16 which is equal to -18.
the minimum point of the second equation is double the minimum point of the first equation.
the graph of both equations is shown below:
you can see hey have the same roots (x-intercepts) and the minimum value of the first equation is double the minimum value of the second equation.
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