We know that
The Standard Form<span> of a </span>Quadratic Equation<span> is
</span> ax²<span> + bx + c = 0
</span>
then
for x=0 y=-6
a*0² + b*0 + c = -6---------> c=-6
for x=1 y=2
a*1² + b*1 -6 = 2---------> a+b=8---------> a=8-b---------> equation 1
for x=3 y=-0
a*3² + b*3 -6 = 0---------> 9a+3b=6---------> equation 2
<span>substituting 1 in 2
9*(8-b)+3b=6------> 72-9b+3b=6---------> 6b=66--------> b=11
a=8-11-------> a=-3
so
a=-3
b=11
c=-6
</span>Standard Form of a Quadratic Equation is
-3x² + 11x - 6 = 0
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the answer is
</span> -3x² + 11x - 6 = 0<span>
</span>
A comparison between a function and its inverse would show that the domain and range of the original function swap. The domain of the function becomes the range of the inverse, the range of the function becomes the domain of its inverse.
Looking at ordered pairs of the function and its inverse would look like this:
(2,4) on the original function becomes (4,2) on the inverse.
While the graph of a function and its inverse are noticeably different an important thing to note is that it is merely a reflection across the line y=x.
So even though they appear different you are looking at the same relationship just as y vs. x instead of x vs. y
B, short explanation remember the sets are set up like (x,y) in a set there should be no matching x points so if it matches it's not a function :)
long explanation: these are (x,y) sets {(0,3), (3,0), (0,4), (4,0)} the first choice shows that there are two X solutions <span>if they are equal to 0 (0,3) (0,4) so this cant be a function.
second choice</span><span> {(0,2), (2,0), (4,6), (6,4)} does not have two X solutions so this is a function
third choice</span>{(2,6), (3,6), (4,6), (2,0)} has two X solutions if x is equal to 2
and the last choice <span>{(6,2), (2,0), (4,6), (6,4)} has two X solutions if x is equal to 6
hope i helped! if i did please give me brainlest :)
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