F(x) = 9x2 - 4 Find the zeros of the function
1 answer:
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<span>Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?
we have that
y=2x
y=x</span>²+2
y=2x²
using a graph tool
see the attached figure
<span>We can affirm the following
</span>the three graphs present the same domain-----> the interval (-∞,∞)
The range of the graph y=2x is the interval (-∞,∞)
The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)
therefore
<span>Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range</span>
<span>Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?
we have that
y=2x
y=x</span>²+2
y=2x²
using a graph tool
see the attached figure
<span>We can affirm the following
</span>the three graphs present the same domain-----> the interval (-∞,∞)
The range of the graph y=2x is the interval (-∞,∞)
The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)
therefore
<span>Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range</span>
Answer:
A
Step-by-step explanation:
if (x1,y1) and (x2,y2) are the extremities of diameter,then eq. of circle is
(x-x1)(x-x2)+(y-y1)(y-y2)=0
reqd. eq. is (x+1)(x-5)+(y+9)(y-1)=0
center is (2,-4)
r=√(2²+(-4)²-(-14))
=√(4+16+14)
=√(34)
eq. of circle is (x-2)²+(y+4)²=34
or
(x²-4x)+(y²+8y)=14
(x²-4x+4)+(y²+8y+16)=14+4+16
(x-2)²+(y+4)²=34