You can give random values such as <span> (1,2.5) (2,9) (-1,2.5) (-2,9) and it is a parabola. You'll see that the range is (-infinte, 0] and the domain is (-infinit, infinite). </span>I hope that this is the answer that you were looking for and it has helped you.
<h3>
Answer: False</h3>
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Explanation:
I'm assuming you meant to type out
(y-2)^2 = y^2-6y+4
This equation is not true for all real numbers because the left hand side expands out like so
(y-2)^2
(y-2)(y-2)
x(y-2) .... let x = y-2
xy-2x
y(x)-2(x)
y(y-2)-2(y-2) ... replace x with y-2
y^2-2y-2y+4
y^2-4y+4
So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4
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Another approach is to pick some y value such as y = 2 to find that
(y-2)^2 = y^2-6y+4
(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2
0^2 = 2^2 - 6(2) + 4
0 = 4 - 6(2) + 4
0 = 4 - 12 + 4
0 = -4
We get a false statement. This is one counterexample showing the given equation is not true for all values of y.
9514 1404 393
Answer:
q = 40
Step-by-step explanation:
When the quadratic has roots p and r, it can be factored as ...
(x -p)(x -r) = x² -(p+r)x +pr
So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...
p + r = 14
p - r = 6
2p = 20 . . . add the two equations
p = 10
r = 14 -p = 4
The value of interest is then ...
q = pr = (10)(4)
q = 40
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The graph shows the roots to be 4 and 10, as we found.
Answer:
depends on how you understand math.
