Question 2167: What is the vertex and X-intercepts of the parabola -14x^2+36-30=0 , and how do you know?
Answer by matthew_sessoms(39)   (Show Source):
You can put this solution on YOUR website! I believe you mean -14x^2 + 36x - 30 = 0. Right? I believe you're the one who asked about the domain & range.
You can find the vertex and x-intercepts either by "graphing" or by the "complete the square" (CTS) method. I will do both.
CTS states that y=a(x-h)^2+k, where (h,k) is the vertex. Noticed the sign of "h" changed. Always change the sign of h, however, keep the sign of k. This is how you complete the square.
First, you have to get the x^2 coefficient to be 1. To do this we must factor out the -14 from the x terms only (not the -30).
-14(x^2 - 18/7x)-30=0 (36/-14 equals -18/7)
Now, we have to take half (or divide by 2) of the -18/7 (which is -9/7) and then square that (which is 81/49). Put the 81/49 in the parentheses, like this:
-14(x^2 - 18/7x + 81/49) - 30 = 0
Third we have to multiply 81/49 by the -14 that is factored out (which is -162/7) and then change the sign and put it beside the -30, like this:
-14(x^2 - 18/7x + 81/49) + 162/7 - 30 = 0
Fourth, we have to factor the equation inside the parentheses. Since we have "completed the square", this is real easy to factor. Just take half of the -18/7 (which is -9/7) and replace this with the -18/7 and then reduce the x^2 to x and square the parentheses. As well as combined the 162/7 - 30, like this:
-14(x - 9/7)^2 - 48/7 = 0
This is now in the form of y = a(x-h)^2 + k, where (h,k) is the vertex. So the vertex is (9/7, -48/7). This vertex is the minimum because "a" (which is -14) is a negative. If it were a positive it would be the maximum.
To find the x-intercepts, you can graph or simply solve for x (either by factoring or using the quadratic equation). Since this equation cannot be factored, we must use the quadratic formula: 
Plug it in, like this  , which simplifies to be 
I don't know if you know anything about "imaginary/complex numbers" but we will have one. Since we cannot take the square root of a negative number, there is NO real x-intercepts, only imaginary/complex x-intercepts. This means it doesn't cross the x-axis at all!
To find the imaginary/complex intercepts, we can simply the equation more, like this: 
So the vertex is (9/7, -48/7) and the x-intercept is
Here is the graph to prove my statements:
|
|
|
